# 4.5 First-order Linear Equations – Calculus Volume 2 | OpenStax

### Learning Objectives

• 4.5.1
Write a first-order linear differential gear equation in standard mannequin .
• 4.5.2
Find an desegregate factor and use it to solve a first-order analogue differential equation.

• 4.5.3
Solve applied problems involving first-order linear differential equations .

earlier, we studied an application of a first-order differential equation that involved solving for the speed of an object. In particular, if a testis is thrown up with an initial speed of v0v0 ft/s, then an initial-value problem that describes the speed of the ball after terrestrial time seconds is given by
dvdt=−32, v ( 0 ) =v0.dvdt=−32, five ( 0 ) =v0 .
This model assumes that the only impel acting on the ball is gravity. now we add to the trouble by allowing for the possibility of air underground acting on the ball .
Air resistance always acts in the direction opposite to gesture. therefore if an object is rising, air immunity acts in a down direction. If the object is falling, air resistance acts in an up focus ( Figure 4.24 ). There is no exact relationship between the speed of an object and the air out resistance acting on it. For very little objects, tune resistance is proportional to speed ; that is, the wedge due to air travel resistance is numerically adequate to some constant kk times v.v. For larger ( for example, baseball-sized ) objects, depending on the shape, air travel resistance can be approximately proportional to the square of the speed. In fact, tune underground may be proportional to v1.5, v1.5, or v0.9, v0.9, or some other might of v.v .

Figure

4.24

Forces acting on a moving baseball: gravity acts in a downward direction and air resistance acts in a direction opposite to the direction of motion.

We will work with the linear approximation for atmosphere resistance. If we assume k > 0, k > 0, then the construction for the force FAFA due to air resistance is given by FA=−kv.FA=−kv. Therefore the summarize of the forces acting on the object is equal to the sum of the gravitational military unit and the force due to air resistance. This, in turn, is peer to the mass of the object multiplied by its acceleration at time micronesia ( Newton ’ s second law ). This gives us the differential equality
mdvdt=−kv−mg.mdvdt=−kv−mg .
finally, we impose an initial condition vanadium ( 0 ) =v0, v ( 0 ) =v0, where v0v0 is the initial speed measured in meters per second. This makes g=9.8m/s2.g=9.8m/s2. The initial-value trouble becomes
mdvdt=−kv−mg,v(0)=v0.mdvdt=−kv−mg,v(0)=v0.

(4.13)

The derived function equality in this initial-value problem is an example of a first-order linear derived function equation. ( Recall that a differential equation is first-order if the highest-order derivative instrument that appears in the equation is 1. ) 1. ) In this section, we study first-order analogue equations and examine a method for finding a general solution to these types of equations, a well as solving initial-value problems involving them .

### definition

A first-order differential equation is linear if it can be written in the class
a(x)y′+b(x)y=c(x),a(x)y′+b(x)y=c(x),

(4.14)

where a ( x ), boron ( ten ), a ( x ), barn ( x ), and c ( x ) hundred ( adam ) are arbitrary functions of x.x .
Remember that the strange function yy depends on the variable ten ; x ; that is, xx is the mugwump varying and yy is the dependent varying. Some examples of first-order linear differential gear equations are
( 3×2−4 ) y′+ ( x−3 ) y=sinx ( sinx ) y′− ( cosx ) y=cotx4xy′+ ( 3lnx ) y=x3−4x. ( 3×2−4 ) y′+ ( x−3 ) y=sinx ( sinx ) y′− ( cosx ) y=cotx4xy′+ ( 3lnx ) y=x3−4x .
Examples of first-order nonlinear differential equations include
( y′ ) 4− ( y′ ) 3= ( 3x−2 ) ( y+4 ) 4y′+3y3=4x−5 ( y′ ) 2=siny+cosx. ( y′ ) 4− ( y′ ) 3= ( 3x−2 ) ( y+4 ) 4y′+3y3=4x−5 ( y′ ) 2=siny+cosx .
These equations are nonlinear because of terms like ( y′ ) 4, y3, ( y′ ) 4, y3, etc. due to these terms, it is impossible to put these equations into the lapp form as equation 4.14 .

### Standard Form

Consider the differential equation
( 3×2−4 ) y′+ ( x−3 ) y=sinx. ( 3×2−4 ) y′+ ( x−3 ) y=sinx .
Our main goal in this section is to derive a solution method acting for equations of this form. It is useful to have the coefficient of y′y′ be equal to 1.1. To make this find, we divide both sides by 3×2−4.3×2−4 .
y′+ ( x−33×2−4 ) y=sinx3x2−4y′+ ( x−33×2−4 ) y=sinx3x2−4
This is called the standard form of the derived function equation. We will use it late when finding the solution to a general first-order linear differential equality. Returning to Equation 4.14, we can divide both sides of the equation by a ( adam ) .a ( x ). This leads to the equation
y′+b(x)a(x)y=c(x)a(x).y′+b(x)a(x)y=c(x)a(x).

(4.15)

now define phosphorus ( x ) =b ( x ) a ( x ) p ( x ) =b ( x ) a ( x ) and q ( x ) =c ( x ) a ( x ) .q ( adam ) =c ( x ) a ( ten ). then Equation 4.14 becomes
y′+p(x)y=q(x).y′+p(x)y=q(x).

(4.16)

We can write any first-order analogue differential equality in this class, and this is referred to as the standard form for a first-order analogue differential equation .

### exemplar 4.15

#### Writing First-Order Linear Equations in Standard Form

Put each of the take after first-order linear derived function equations into standard form. Identify phosphorus ( x ) p ( x ) and q ( x ) q ( x ) for each equality .

1. y′=3x−4yy′=3x−4y
2. 3xy′4y−3=23xy′4y−3=2 (here x≠0)x≠0)
3. y=3y′−4×2+5y=3y′−4×2+5

#### solution

1. Add 4y4y to both sides:
y′+4y=3x.y′+4y=3x.
In this equation, p(x)=4p(x)=4 and q(x)=3x.q(x)=3x.
2. Multiply both sides by 4y−3,4y−3, then subtract 8y8y from each side:
3xy′4y−3=23xy′=2 ( 4y−3 ) 3xy′=8y−63xy′−8y=−6.3xy′4y−3=23xy′=2 ( 4y−3 ) 3xy′=8y−63xy′−8y=−6.
Finally, divide both sides by 3x3x to make the coefficient of y′y′ equal to 1:1:
y′−83xy=−2x.y′−83xy=−2x.

(4.17)

x≠0.x≠0. (If x=0x=0 then the original equation becomes 0=2,0=2, which is clearly a false statement.)
In this equation, p(x)=−83xp(x)=−83x and q(x)=−23x.q(x)=−23x.

3. Subtract yy from each side and add 4×2−5:4×2−5:
3y′−y=4×2−5.3y′−y=4×2−5.
Next divide both sides by 3:3:
y′−13y=43×2−53.y′−13y=43×2−53.
In this equation, p(x)=−13p(x)=−13 and q(x)=43×2−53.q(x)=43×2−53.

### checkpoint 4.15

Put the equation ( x+3 ) y′2x−3y−4=5 ( x+3 ) y′2x−3y−4=5 into standard human body and identify p ( x ) p ( x ) and q ( x ) .q ( x ) .

### Integrating Factors

We now develop a solution proficiency for any first-order linear differential gear equation. We start with the standard form of a first-order linear differential gear equality :
y′+p(x)y=q(x).y′+p(x)y=q(x).

(4.18)

The beginning terminus on the left side of Equation 4.15 is the derivative instrument of the unknown function, and the second base term is the product of a known officiate with the unknown function. This is reasonably evocative of the product predominate from the Differentiation Rules section. If we multiply equality 4.16 by a yet-to-be-determined function μ ( x ), μ ( x ), then the equation becomes
μ(x)y′+μ(x)p(x)y=μ(x)q(x).μ(x)y′+μ(x)p(x)y=μ(x)q(x).

(4.19)

The left slope Equation 4.18 can be matched absolutely to the product rule :
ddx [ farad ( x ) g ( x ) ] =f′ ( x ) gigabyte ( x ) +f ( x ) g′ ( x ) .ddx [ f ( x ) thousand ( x ) ] =f′ ( x ) gram ( x ) +f ( x ) g′ ( x ) .
Matching condition by term gives y=f ( adam ), g ( x ) =μ ( x ), y=f ( ten ), g ( ten ) =μ ( x ), and g′ ( x ) =μ ( x ) phosphorus ( x ) .g′ ( x ) =μ ( x ) p ( x ). Taking the derivative of gigabyte ( x ) =μ ( x ) thousand ( x ) =μ ( x ) and setting it equal to the right-hand side of g′ ( x ) =μ ( x ) phosphorus ( x ) g′ ( x ) =μ ( x ) phosphorus ( x ) leads to
μ′ ( x ) =μ ( x ) phosphorus ( x ) .μ′ ( x ) =μ ( x ) p ( x ) .
This is a first-order, dissociable differential equation for μ ( x ) .μ ( x ). We know phosphorus ( x ) phosphorus ( ten ) because it appears in the differential gear equation we are solving. Separating variables and integrating yields
μ′ ( x ) μ ( x ) =p ( x ) ∫μ′ ( x ) μ ( x ) dx=∫p ( x ) dxln|μ ( x ) |=∫p ( x ) dx+Celn|μ ( x ) |=e∫p ( x ) dx+C|μ ( x ) |=C1e∫p ( x ) dxμ ( x ) =C2e∫p ( x ) dx.μ′ ( x ) μ ( x ) =p ( x ) ∫μ′ ( x ) μ ( x ) dx=∫p ( x ) dxln|μ ( x ) |=∫p ( x ) dx+Celn|μ ( x ) |=e∫p ( x ) dx+C|μ ( x ) |=C1e∫p ( x ) dxμ ( x ) =C2e∫p ( x ) dx .
here C2C2 can be an arbitrary ( positive or negative ) constant. This leads to a general method acting for solving a first-order linear differential equality. We first gear multiply both sides of Equation 4.16 by the integration component μ ( x ) .μ ( x ). This gives
μ(x)y′+μ(x)p(x)y=μ(x)q(x).μ(x)y′+μ(x)p(x)y=μ(x)q(x).

(4.20)

The left-hand side of Equation 4.20 can be rewritten as ddx ( μ ( x ) y ) .ddx ( μ ( x ) yttrium ) .
ddx(μ(x)y)=μ(x)q(x).ddx(μ(x)y)=μ(x)q(x).

(4.21)

following integrate both sides of Equation 4.21 with respect to x.x .
∫ddx(μ(x)y)dx=∫μ(x)q(x)dxμ(x)y=∫μ(x)q(x)dx.∫ddx(μ(x)y)dx=∫μ(x)q(x)dxμ(x)y=∫μ(x)q(x)dx.

(4.22)

Divide both sides of Equation 4.22 by μ ( adam ) : μ ( ten ) :
y=1μ(x)[∫μ(x)q(x)dx+C].y=1μ(x)[∫μ(x)q(x)dx+C].

(4.23)

Since μ ( x ) μ ( x ) was previously calculated, we are now finished. An important note about the integrate changeless C : speed of light : It may seem that we are inconsistent in the usage of the integration changeless. however, the integral involving phosphorus ( x ) p ( x ) is necessary in ordering to find an integrate gene for Equation 4.15. only one integrate gene is needed in order to solve the equation ; therefore, it is safe to assign a value for CC for this built-in. We chose C=0.C=0. When calculating the integral inside the brackets in Equation 4.21, it is necessary to keep our options open for the value of the integrate constant, because our goal is to find a general kin of solutions to Equation 4.15. This integration factor guarantees just that .

### Problem-Solving Strategy

#### Problem-Solving strategy : Solving a First-order Linear Differential Equation

1. Put the equation into standard form and identify p(x)p(x) and q(x).q(x).
2. Calculate the integrating factor μ(x)=e∫p(x)dx.μ(x)=e∫p(x)dx.
3. Multiply both sides of the differential equation by μ(x).μ(x).
4. Integrate both sides of the equation obtained in step 3,3, and divide both sides by μ(x).μ(x).
5. If there is an initial condition, determine the value of C.C.

### exercise 4.16

#### Solving a First-order Linear Equation

Find a general solution for the differential equation xy′+3y=4×2−3x.xy′+3y=4×2−3x. Assume x > 0.x > 0 .

#### solution

1. To put this differential equation into standard form, divide both sides by x:x:
y′+3xy=4x−3.y′+3xy=4x−3.
Therefore p(x)=3xp(x)=3x and q(x)=4x−3.q(x)=4x−3.
2. The integrating factor is μ(x)=e∫(3/x)dx=e3lnx=x3.μ(x)=e∫(3/x)dx=e3lnx=x3.
3. Multiplying both sides of the differential equation by μ(x)μ(x) gives us
x3y′+x3 ( 3x ) y=x3 ( 4x−3 ) x3y′+3x2y=4×4−3x3ddx ( x3y ) =4×4−3×3.x3y′+x3 ( 3x ) y=x3 ( 4x−3 ) x3y′+3x2y=4×4−3x3ddx ( x3y ) =4×4−3×3 .
4. Integrate both sides of the equation.
∫ddx ( x3y ) dx=∫4×4−3x3dxx3y=4×55−3×44+Cy=4×25−3×4+Cx−3.∫ddx ( x3y ) dx=∫4×4−3x3dxx3y=4×55−3×44+Cy=4×25−3×4+Cx−3 .
5. There is no initial value, so the problem is complete.

#### analysis

You may have noticed the circumstance that was imposed on the derived function equality ; namely, x > 0.x > 0. For any nonzero respect of C, C, the general solution is not defined at x=0.x=0. Furthermore, when ten < 0, x < 0, the integrating factor changes. The integration factor is given by Equation 4.19 as degree fahrenheit ( x ) =e∫p ( x ) dx.f ( x ) =e∫p ( x ) dx. For this p ( x ) p ( adam ) we get e∫p ( x ) dx=e∫ ( 3/x ) dx=e3ln|x|=|x|3, e∫p ( x ) dx=e∫ ( 3/x ) dx=e3ln|x|=|x|3 , since x < 0.x < 0. The behavior of the general solution changes at x=0x=0 largely due to the fact that p ( x ) phosphorus ( x ) is not defined there .

### checkpoint 4.16

Find the general solution to the differential equation ( x−2 ) y′+y=3×2+2x. ( x−2 ) y′+y=3×2+2x. Assume x > 2.x > 2 .
now we use the same scheme to find the solution to an initial-value problem .

### example 4.17

#### A First-order Linear Initial-Value Problem

Solve the initial-value problem
y′+3y=2x−1, yttrium ( 0 ) =3.y′+3y=2x−1, y ( 0 ) =3 .

#### solution

1. This differential equation is already in standard form with p(x)=3p(x)=3 and q(x)=2x−1.q(x)=2x−1.
2. The integrating factor is μ(x)=e∫3dx=e3x.μ(x)=e∫3dx=e3x.
3. Multiplying both sides of the differential equation by μ(x)μ(x) gives
e3xy′+3e3xy= ( 2x−1 ) e3xddx [ ye3x ] = ( 2x−1 ) e3x.e3xy′+3e3xy= ( 2x−1 ) e3xddx [ ye3x ] = ( 2x−1 ) e3x.
Integrate both sides of the equation:
∫ddx [ ye3x ] dx=∫ ( 2x−1 ) e3xdxye3x=e3x3 ( 2x−1 ) −∫23e3xdxye3x=e3x ( 2x−1 ) 3−2e3x9+Cy=2x−13−29+Ce−3xy=2×3−59+Ce−3x.∫ddx [ ye3x ] dx=∫ ( 2x−1 ) e3xdxye3x=e3x3 ( 2x−1 ) −∫23e3xdxye3x=e3x ( 2x−1 ) 3−2e3x9+Cy=2x−13−29+Ce−3xy=2×3−59+Ce−3x .
4. Now substitute x=0x=0 and y=3y=3 into the general solution and solve for C:C:
y=23x−59+Ce−3×3=23 ( 0 ) −59+Ce−3 ( 0 ) 3=−59+CC=329.y=23x−59+Ce−3×3=23 ( 0 ) −59+Ce−3 ( 0 ) 3=−59+CC=329.
Therefore the solution to the initial-value problem is
y=23x−59+329e−3x.y=23x−59+329e−3x .

### checkpoint 4.17

Solve the initial-value problem y′−2y=4x+3y ( 0 ) =−2.y′−2y=4x+3y ( 0 ) =−2 .

### Applications of First-order Linear Differential Equations

We look at two different applications of first-order linear differential gear equations. The first involves air resistance as it relates to objects that are rising or falling ; the second involves an electric circuit. other applications are numerous, but most are solved in a similar fashion .

#### Free fall with air resistance

We discussed air resistance at the begin of this section. The next example shows how to apply this concept for a ball in vertical motion. other factors can affect the pull of air resistance, such as the size and form of the object, but we ignore them here .

### exemplar 4.18

#### A Ball with Air Resistance

A racquetball is hit straight up with an initial speed of 22 m/s. The mass of a racquetball is approximately 0.04270.0427 kg. Air underground acts on the ball with a storm numerically peer to 0.5v,0.5v, where vv represents the speed of the ball at time t.t .

1. Find the velocity of the ball as a function of time.
2. How long does it take for the ball to reach its maximum height?
3. If the ball is hit from an initial height of 11 meter, how high will it reach?

#### solution

1. The mass m=0.0427kg,k=0.5,m=0.0427kg,k=0.5, and g=9.8m/s2.g=9.8m/s2. The initial velocity is v0=2v0=2 m/s. Therefore the initial-value problem is
0.0427dvdt=−0.5v−0.0427 ( 9.8 ), v0=2.0.0427dvdt=−0.5v−0.0427 ( 9.8 ), v0=2.
Dividing the differential equation by 0.04270.0427 gives
dvdt=−11.7096v−9.8, v0=2.dvdt=−11.7096v−9.8, v0=2.
The differential equation is linear. Using the problem-solving strategy for linear differential equations:
Step 1. Rewrite the differential equation as dvdt+11.7096v=−9.8.dvdt+11.7096v=−9.8. This gives p(t)=11.7096p(t)=11.7096 and q(t)=−9.8q(t)=−9.8
Step 2. The integrating factor is μ(t)=e∫11.7096dt=e11.7096t.μ(t)=e∫11.7096dt=e11.7096t.
Step 3. Multiply the differential equation by μ(t):μ(t):
e11.7096tdvdt+11.7096ve11.7096t=−9.8e11.7096tddt [ ve11.7096t ] =−9.8e11.7096t.e11.7096tdvdt+11.7096ve11.7096t=−9.8e11.7096tddt [ ve11.7096t ] =−9.8e11.7096t.
Step 4. Integrate both sides:
∫ddt [ ve11.7096t ] dt=∫−9.8e11.7096tdtve11.7096t=−9.811.7096e11.7096t+Cv ( deoxythymidine monophosphate ) =−0.8369+Ce−11.7096t.∫ddt [ ve11.7096t ] dt=∫−9.8e11.7096tdtve11.7096t=−9.811.7096e11.7096t+Cv ( deoxythymidine monophosphate ) =−0.8369+Ce−11.7096t.
Step 5. Solve for CC using the initial condition v0=v(0)=2:v0=v(0)=2:
vanadium ( thymine ) =−0.8369+Ce−11.7096tv ( 0 ) =−0.8369+Ce−11.7096 ( 0 ) 2=−0.8369+CC=2.8369.v ( deoxythymidine monophosphate ) =−0.8369+Ce−11.7096tv ( 0 ) =−0.8369+Ce−11.7096 ( 0 ) 2=−0.8369+CC=2.8369.
Therefore the solution to the initial-value problem is v(t)=2.8369e−11.7096t−0.8369.v(t)=2.8369e−11.7096t−0.8369.
2. The ball reaches its maximum height when the velocity is equal to zero. The reason is that when the velocity is positive, it is rising, and when it is negative, it is falling. Therefore when it is zero, it is neither rising nor falling, and is at its maximum height:
2.8369e−11.7096t−0.8369=02.8369e−11.7096t=0.8369e−11.7096t=0.83692.8369≈0.295lne−11.7096t=ln0.295≈−1.221−11.7096t=−1.221t≈0.104.2.8369e−11.7096t−0.8369=02.8369e−11.7096t=0.8369e−11.7096t=0.83692.8369≈0.295lne−11.7096t=ln0.295≈−1.221−11.7096t=−1.221t≈0.104.
Therefore it takes approximately 0.1040.104 second to reach maximum height.
3. To find the height of the ball as a function of time, use the fact that the derivative of position is velocity, i.e., if h(t)h(t) represents the height at time t,t, then h′(t)=v(t).h′(t)=v(t). Because we know v(t)v(t) and the initial height, we can form an initial-value problem:
h′ ( triiodothyronine ) =2.8369e−11.7096t−0.8369, h ( 0 ) =1.h′ ( thyroxine ) =2.8369e−11.7096t−0.8369, planck’s constant ( 0 ) =1.
Integrating both sides of the differential equation with respect to tt gives
∫h′ ( triiodothyronine ) dt=∫2.8369e−11.7096t−0.8369dth ( deoxythymidine monophosphate ) =−2.836911.7096e−11.7096t−0.8369t+Ch ( t ) =−0.2423e−11.7096t−0.8369t+C.∫h′ ( metric ton ) dt=∫2.8369e−11.7096t−0.8369dth ( deoxythymidine monophosphate ) =−2.836911.7096e−11.7096t−0.8369t+Ch ( deoxythymidine monophosphate ) =−0.2423e−11.7096t−0.8369t+C.
Solve for CC by using the initial condition:
heat content ( thymine ) =−0.2423e−11.7096t−0.8369t+Ch ( 0 ) =−0.2423e−11.7096 ( 0 ) −0.8369 ( 0 ) +C1=−0.2423+CC=1.2423.h ( deoxythymidine monophosphate ) =−0.2423e−11.7096t−0.8369t+Ch ( 0 ) =−0.2423e−11.7096 ( 0 ) −0.8369 ( 0 ) +C1=−0.2423+CC=1.2423.
Therefore
hydrogen ( metric ton ) =−0.2423e−11.7096t−0.8369t+1.2423.h ( metric ton ) =−0.2423e−11.7096t−0.8369t+1.2423.
After 0.1040.104 second, the height is given by
h(0.104)=−0.2423e−11.7096t−0.8369t+1.2423≈1.0836h(0.104)=−0.2423e−11.7096t−0.8369t+1.2423≈1.0836 meter.

### checkpoint 4.18

The burden of a penny is 2.52.5 grams ( United States Mint, “ Coin Specifications, ” accessed April 9, 2015, hypertext transfer protocol : //www.usmint.gov/about_the_mint/ ? action=coin_specifications ), and the upper berth notice deck of the Empire State Building is 369369 meters above the street. Since the penny is a little and relatively polish object, air resistance acting on the penny is actually quite small. We assume the air travel immunity is numerically adequate to 0.0025v.0.0025v. furthermore, the penny is dropped with no initial speed imparted to it .

1. Set up an initial-value problem that represents the falling penny.
2. Solve the problem for v(t).v(t).
3. What is the terminal velocity of the penny (i.e., calculate the limit of the velocity as tt approaches infinity)?

#### Electrical Circuits

A source of electromotive force ( for example, a battery or generator ) produces a flow of current in a close circuit, and this stream produces a electric potential drop across each resistor, inductor, and capacitor in the lap. Kirchhoff ’ south Loop Rule states that the total of the electric potential drops across resistors, inductors, and capacitors is equal to the total electromotive force in a close up racing circuit. We have the following three results :

1. The voltage drop across a resistor is given by
ER=Ri, ER=Ri,
where RR is a constant of proportionality called the resistance, and ii is the current.
2. The voltage drop across an inductor is given by
EL=Li′, EL=Li′,
where LL is a constant of proportionality called the inductance, and ii again denotes the current.
3. The voltage drop across a capacitor is given by
EC=1Cq, EC=1Cq ,

where CC is a constant of proportionality called the capacitance, and qq is the instantaneous charge on the capacitor. The kinship between two and qq is i=q′.i=q′ .
We use units of volts ( V ) ( V ) to measure voltage E, E, amperes ( A ) ( A ) to measure current iodine, iodine, coulomb ( C ) ( C ) to measure charge q, q, ohm ( Ω ) ( Ω ) to measure resistance R, R, henry ( H ) ( H ) to measure inductor L, L, and farads ( F ) ( F ) to measure capacitance C.C. Consider the racing circuit in Figure 4.25 .

Figure

4.25

A typical electric circuit, containing a voltage generator

(

V
S

)

,

(

V
S

)

,

capacitor

(
C
)

,

(
C
)

,

inductor

(
L
)

,

(
L
)

,

and resistor

(
R
)

.

(
R
)

.

Applying Kirchhoff ’ second Loop Rule to this circuit, we let EE denote the electromotive force supplied by the voltage generator. then
EL+ER+EC=E.EL+ER+EC=E .
Substituting the expressions for EL, ER, EL, ER, and ECEC into this equation, we obtain
Li′+Ri+1Cq=E.Li′+Ri+1Cq=E.

(4.24)

If there is no capacitor in the circuit, then the equation becomes
Li′+Ri=E.Li′+Ri=E.

(4.25)

This is a first-order differential equality in i.i. The tour is referred to as an LRLR circuit .
adjacent, speculate there is no inductor in the tour, but there is a capacitor and a resistor, so L=0, R≠0, L=0, R≠0, and C≠0.C≠0. then Equation 4.23 can be rewritten as
Rq′+1Cq=E,Rq′+1Cq=E,

(4.26)

which is a first-order analogue differential gear equation. This is referred to as an RC circuit. In either encase, we can set up and solve an initial-value trouble .

### example 4.19

#### Finding Current in an RL Electric Circuit

A racing circuit has in serial an electromotive force out given by E=50sin20tV, E=50sin20tV, a resistor of 5Ω,5Ω, and an inductor of 0.4H.0.4H. If the initial current is 0,0, find the stream at time triiodothyronine > 0.t > 0 .

#### solution

We have a resistor and an inductor in the tour, so we use equality 4.24. The voltage drop across the resistor is given by ER=Ri=5i.ER=Ri=5i. The voltage drop across the inductor is given by EL=Li′=0.4i′.EL=Li′=0.4i′. The electromotive impel becomes the right side of Equation 4.24. Therefore equation 4.24 becomes
0.4 one ′ + 5 iodine = 50 sin 20 t. 0.4 i ′ + 5 one = 50 sin 20 t .
Dividing both sides by 0.40.4 gives the equality
one ′ + 12.5 one = 125 sin 20 deoxythymidine monophosphate. i ′ + 12.5 iodine = 125 sin 20 metric ton .
Since the initial current is 0, this leave gives an initial stipulate of one ( 0 ) =0.i ( 0 ) =0. We can solve this initial-value problem using the five-step strategy for solving first-order differential equations .
step 1. Rewrite the differential equation as i′+12.5i=125sin20t.i′+12.5i=125sin20t. This gives phosphorus ( triiodothyronine ) =12.5p ( metric ton ) =12.5 and q ( thymine ) =125sin20t.q ( t ) =125sin20t .
step 2. The integration factor is μ ( thymine ) =e∫12.5dt=e12.5t.μ ( thyroxine ) =e∫12.5dt=e12.5t .
step 3. Multiply the differential gear equation by μ ( deoxythymidine monophosphate ) : μ ( thyroxine ) :
e 12.5 metric ton iodine ′ + 12.5 einsteinium 12.5 thyroxine i = 125 einsteinium 12.5 deoxythymidine monophosphate sin 20 t five hundred five hundred t [ one e 12.5 thymine ] = 125 e 12.5 thymine sin 20 triiodothyronine. e 12.5 thyroxine one ′ + 12.5 east 12.5 thymine one = 125 e 12.5 thyroxine sin 20 triiodothyronine d five hundred t [ one e 12.5 metric ton ] = 125 e 12.5 t sin 20 t .
dance step 4. Integrate both sides :
∫ five hundred five hundred t [ i e 12.5 deoxythymidine monophosphate ] vitamin d t = ∫ 125 e 12.5 metric ton sin 20 t vitamin d thymine one e 12.5 t = ( 250 drop the ball 20 thyroxine − 400 cos 20 metric ton 89 ) vitamin e 12.5 t + C iodine ( metric ton ) = 250 sin 20 thyroxine − 400 cos 20 t 89 + C east −12.5 deoxythymidine monophosphate. ∫ d d t [ one e 12.5 thymine ] five hundred deoxythymidine monophosphate = ∫ 125 e 12.5 thymine sin 20 deoxythymidine monophosphate five hundred triiodothyronine one e 12.5 thymine = ( 250 sin 20 triiodothyronine − 400 cos 20 metric ton 89 ) e 12.5 thyroxine + C iodine ( metric ton ) = 250 sin 20 t − 400 cos 20 t 89 + C e −12.5 t .
step 5. Solve for CC using the initial condition five ( 0 ) =2 : five ( 0 ) =2 :
iodine ( deoxythymidine monophosphate ) = 250 sin 20 thyroxine − 400 cos 20 metric ton 89 + C einsteinium −12.5 triiodothyronine i ( 0 ) = 250 sin 20 ( 0 ) − 400 cos 20 ( 0 ) 89 + C einsteinium −12.5 ( 0 ) 0 = − 400 89 + C C = 400 89. one ( deoxythymidine monophosphate ) = 250 sin 20 deoxythymidine monophosphate − 400 cos 20 thyroxine 89 + C vitamin e −12.5 t one ( 0 ) = 250 sin 20 ( 0 ) − 400 cos 20 ( 0 ) 89 + C vitamin e −12.5 ( 0 ) 0 = − 400 89 + C C = 400 89 .
therefore the solution to the initial-value trouble is one ( triiodothyronine ) =250sin20t−400cos20t+400e−12.5t89=250sin20t−400cos20t89+400e−12.5t89.i ( t ) =250sin20t−400cos20t+400e−12.5t89=250sin20t−400cos20t89+400e−12.5t89 .
The inaugural term can be rewritten as a individual cosine affair. First, breed and separate by 2502+4002=5089:2502+4002=5089 :
250 sin 20 metric ton − 400 cos 20 thyroxine 89 = 50 89 89 ( 250 sin 20 thyroxine − 400 cos 20 t 50 89 ) = − 50 89 89 ( 8 carbon monoxide 20 deoxythymidine monophosphate 89 − 5 drop the ball 20 deoxythymidine monophosphate 89 ). 250 sine 20 thymine − 400 cos 20 triiodothyronine 89 = 50 89 89 ( 250 sin 20 thyroxine − 400 cos 20 t 50 89 ) = − 50 89 89 ( 8 cobalt 20 triiodothyronine 89 − 5 sin 20 thyroxine 89 ) .
following, define φφ to be an acuate fish such that cosφ=889.cosφ=889. then sinφ=589sinφ=589 and
− 50 89 89 ( 8 colorado 20 thymine 89 − 5 sin 20 deoxythymidine monophosphate 89 ) = − 50 89 89 ( cosine φ cos 20 thymine − sin φ sin 20 metric ton ) = − 50 89 89 carbon monoxide ( 20 thyroxine + φ ). − 50 89 89 ( 8 carbon monoxide 20 thyroxine 89 − 5 drop the ball 20 thymine 89 ) = − 50 89 89 ( conscientious objector φ cos 20 t − drop the ball φ sin 20 thyroxine ) = − 50 89 89 cobalt ( 20 deoxythymidine monophosphate + φ ) .
consequently the solution can be written as
iodine ( triiodothyronine ) = − 50 89 89 conscientious objector ( 20 thymine + φ ) + 400 vitamin e −12.5 t 89. one ( deoxythymidine monophosphate ) = − 50 89 89 conscientious objector ( 20 t + φ ) + 400 east −12.5 t 89 .
The second term is called the attenuation term, because it disappears quickly as thyroxine grows larger. The phase careen is given by φ, φ, and the amplitude of the steady-state current is given by 508989.508989. The graph of this solution appears in number 4.26 :

Figure

4.26

### checkpoint 4.19

A racing circuit has in series an electromotive force given by E=20sin5tE=20sin5t V, a capacitor with capacitance 0.02F,0.02F, and a resistor of 8Ω.8Ω. If the initial charge is 4C,4C, find the commission at fourth dimension triiodothyronine > 0.t > 0 .

### section 4.5 Exercises

Are the surveil derived function equations linear ? Explain your intelligent .
208 .
d y vitamin d x = x 2 yttrium + sin x d yttrium d x = x 2 yttrium + drop the ball x
209 .
vitamin d y five hundred thymine = thymine yttrium five hundred y vitamin d thyroxine = deoxythymidine monophosphate y
210 .
vitamin d y vitamin d metric ton + y 2 = x vitamin d y five hundred thymine + y 2 = x
211 .
yttrium ′ = x 3 + east x y ′ = x 3 + east ten
212 .
yttrium ′ = y + vitamin e y y ′ = y + e y
Write the comply first-order differential equations in standard mannequin .
213 .
yttrium ′ = x 3 yttrium + sin x y ′ = x 3 yttrium + sin ten
214 .
y ′ + 3 y − ln x = 0 y ′ + 3 yttrium − ln x = 0
215 .
− x y ′ = ( 3 x + 2 ) yttrium + x einsteinium x − x y ′ = ( 3 x + 2 ) y + x vitamin e ten
216 .
five hundred yttrium vitamin d thymine = 4 yttrium + deoxythymidine monophosphate y + tangent t five hundred y d t = 4 yttrium + triiodothyronine yttrium + tan metric ton
217 .
d y five hundred deoxythymidine monophosphate = y x ( x + 1 ) five hundred yttrium vitamin d thymine = yttrium x ( x + 1 )
What are the desegregate factors for the following differential equations ?
218 .
yttrium ′ = x y + 3 yttrium ′ = x y + 3
219 .
yttrium ′ + vitamin e x y = drop the ball x y ′ + e x y = sin ten
220 .
y ′ = x ln ( x ) y + 3 x yttrium ′ = x ln ( x ) y + 3 x
221 .
vitamin d yttrium five hundred x = tanh ( x ) y + 1 five hundred yttrium d x = tanh ( x ) y + 1
222 .
vitamin d y five hundred deoxythymidine monophosphate + 3 thyroxine y = vitamin e triiodothyronine y five hundred yttrium vitamin d metric ton + 3 thymine yttrium = east thyroxine yttrium
Solve the stick to differential equations by using integrating factors .
223 .
yttrium ′ = 3 yttrium + 2 yttrium ′ = 3 y + 2
224 .
yttrium ′ = 2 yttrium − x 2 yttrium ′ = 2 yttrium − x 2
225 .
ten y ′ = 3 y − 6 x 2 ten yttrium ′ = 3 y − 6 x 2
226 .
( ten + 2 ) y ′ = 3 ten + y ( ten + 2 ) yttrium ′ = 3 adam + y
227 .
yttrium ′ = 3 x + x y y ′ = 3 x + x y
228 .
ten yttrium ′ = x + y x y ′ = x + y
229 .
sin ( x ) y ′ = y + 2 ten sin ( x ) y ′ = y + 2 x
230 .
yttrium ′ = y + e x y ′ = y + e ten
231 .
ten y ′ = 3 y + x 2 adam yttrium ′ = 3 y + x 2
232 .
yttrium ′ + ln x = y x y ′ + ln x = y x
Solve the follow derived function equations. Use your calculator to draw a syndicate of solutions. Are there certain initial conditions that change the behavior of the solution ?
233 .
[T] ( x+2 ) y′=2y−1 ( x+2 ) y′=2y−1
234 .
[T] y′=3et/3−2yy′=3et/3−2y
235 .
[T] xy′+y2=sin ( 3t ) xy′+y2=sin ( 3t )
236 .
[T] xy′=2cosxx−3yxy′=2cosxx−3y
237 .
[T] ( x+1 ) y′=3y+x2+2x+1 ( x+1 ) y′=3y+x2+2x+1
238 .
[T] sin ( x ) y′+cos ( x ) y=2xsin ( x ) y′+cos ( x ) y=2x
239 .
[T] x2+1y′=y+2×2+1y′=y+2
240 .
[T] x3y′+2x2y=x+1x3y′+2x2y=x+1
Solve the play along initial-value problems by using integrating factors .
241 .
y ′ + y = x, y ( 0 ) = 3 y ′ + y = x, y ( 0 ) = 3
242 .
y ′ = y + 2 ten 2, y ( 0 ) = 0 y ′ = y + 2 adam 2, yttrium ( 0 ) = 0
243 .
ten yttrium ′ = y − 3 ten 3, yttrium ( 1 ) = 0 x yttrium ′ = y − 3 ten 3, y ( 1 ) = 0
244 .
ten 2 y ′ = x y − ln x, y ( 1 ) = 1 x 2 yttrium ′ = x y − ln x, y ( 1 ) = 1
245 .
( 1 + x 2 ) yttrium ′ = y − 1, y ( 0 ) = 0 ( 1 + x 2 ) y ′ = y − 1, y ( 0 ) = 0
246 .
ten yttrium ′ = y + 2 x ln x, y ( 1 ) = 5 x y ′ = y + 2 x ln x, y ( 1 ) = 5
247 .
( 2 + x ) y ′ = y + 2 + x, y ( 0 ) = 0 ( 2 + x ) y ′ = y + 2 + x, y ( 0 ) = 0
248 .
yttrium ′ = x y + 2 x vitamin e x, y ( 0 ) = 2 y ′ = x y + 2 x vitamin e x, y ( 0 ) = 2
249 .
adam y ′ = y + 2 ten, yttrium ( 0 ) = 1 x y ′ = y + 2 ten, yttrium ( 0 ) = 1
250 .
y ′ = 2 yttrium + x vitamin e x, y ( 0 ) = −1 y ′ = 2 yttrium + x vitamin e x, y ( 0 ) = −1
251 .
A falling object of mass millimeter can reach concluding speed when the embroil effect is proportional to its speed, with proportionality changeless k.k. Set up the differential gear equation and solve for the speed given an initial speed of 0.0 .
252 .
Using your expression from the preceding problem, what is the terminal speed ? ( trace : Examine the limit behavior ; does the speed approach a rate ? )
253 .
[T] Using your equation for concluding speed, solve for the distance fallen. How long does it take to fall 50005000 meters if the mass is 100100 kilograms, the acceleration due to gravity is 9.89.8 m/s2 and the proportionality changeless is 4 ? 4 ?
254 .
A more accurate way to describe end speed is that the scuff force is proportional to the square of speed, with a proportionality constant k.k. Set up the differential equality and solve for the speed .
255 .
Using your saying from the preceding problem, what is the terminal speed ? ( tip : Examine the modification behavior : Does the speed approach a value ? )
256 .
[T] Using your equality for terminal speed, solve for the distance fallen. How long does it take to fall 50005000 meters if the bulk is 100100 kilograms, the acceleration due to gravity is 9.8m/s29.8m/s2 and the proportionality changeless is 4 ? 4 ? Does it take more or less meter than your initial estimate ?
For the come problems, determine how argument associate in arts affects the solution .
257 .
Solve the generic equation y′=ax+y.y′=ax+y. How does varying aa change the demeanor ?
258 .
Solve the generic equation y′=ay+x.y′=ay+x. How does varying aa change the behavior ?
259 .
Solve the generic equation y′=ax+xy.y′=ax+xy. How does varying aa change the behavior ?
260.

Solve the generic equation y′=x+axy.y′=x+axy. How does varying aa change the demeanor ?
261 .
Solve y′−y=ekty′−y=ekt with the initial condition y ( 0 ) =0.y ( 0 ) =0. As kk approaches 1,1, what happens to your formula ?

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