34.5 First Order Linear Differential Equations Problem Set

FIRST ORDER LINEAR DIFFERENTIAL EQUATIONS | PRACTICE PROBLEMS Complete the following to reinforce your understanding of the concept covered in this module.

PROBLEM 1: What is the general solution of the following differential equation? 𝑦 " + 2𝑥𝑦 = 𝑥 '

A. 𝑦 = − 𝐶𝑒 ,-

.

'

.

(

B. 𝑦 = + 𝐶𝑒 ,( '

C. 𝑦 = + 𝐶𝑒 -

.

( '

D. 𝑦 = − 𝐶𝑒 ,-

.

(

SOLUTION 1: We first classify the differential equation to verify which method we should use to solve for the general and particular solutions. We see that the differential equation is a first order linear differential equation.



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The STANDARD FORM OF A FIRST ORDER LINEAR DIFFERENTIAL EQUATION is not provided in the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. We must memorize this formula and understand its application independent of the NCEES Supplied Reference Handbook. The next step in this problem is compare the differential equation with the standard form for a first order linear differential equation. 𝑑𝑦 + 𝑃 𝑥 𝑦 = 𝑄(𝑥) 𝑑𝑥 𝑦 " + 2𝑥𝑦 = 𝑥 After writing the equation in standard form, the 𝑃(𝑥) term can be identified. We see that this equation is already expressed in standard with form the standard terms be represented as: 𝑃 𝑥 = 2𝑥 𝑎𝑛𝑑 𝑄 𝑥 = 𝑥 The FORMULA FOR THE INTEGRATING FACTOR is not provided in the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. We must memorize this formula and understand its application independent of the NCEES Supplied Reference Handbook. Once we have identified the 𝑃 𝑥 term, we can determine the integrating factor by plugging the 𝑃(𝑥) term into the formula for the integrating factor: 𝜇 𝑥 = 𝑒∫ 9



- :-

=𝑒

(- :-

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Multiplying both sides of the differential equation by the integrating factor: 𝜇 𝑥 𝑦 " + 2𝑥𝑦 = 𝜇(𝑥)𝑥 Plugging in the function for the integrating factor: .

.

𝑒 - 𝑦 " + 2𝑥𝑒 - 𝑦 = 𝑥𝑒 -

.

Multiply the equation in standard form by the integrating factor, 𝜇 𝑥 and verify that the left side becomes the product rule (𝜇 𝑥 𝑦 𝑥 )′ and write it as such. The integrating factor is defined so that equation becomes equivalent to: 𝑑 𝜇 𝑥 𝑦 = 𝜇 𝑥 𝑄(𝑥) 𝑑𝑥 𝑑 -. . 𝑒 𝑦′ = 𝑥𝑒 𝑑𝑥 .

.

𝑑 𝑒 - 𝑦′ = 𝑥𝑒 - 𝑑𝑥 We then integrate both sides of the equation with respect to the independent variable, which is “𝑥” in this differential equation. 𝜇 𝑥 𝑦 = ∫ 𝜇 𝑥 𝑄 𝑥 𝑑𝑥 .

.

∫ 𝑑 𝑒 - 𝑦′ = ∫ 𝑥𝑒 - 𝑑𝑥



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Using the integration technique of substitution, we can evaluate the integral of : .

∫ 𝑥𝑒 - 𝑑𝑥 We define our variables of substitution as: 𝑢 = 𝑥( 𝑑𝑢 = 2𝑥 𝑑𝑥 𝑑𝑢 = 𝑥 𝑑𝑥 2 Using these variables, we can re-write our integral as: .

∫ 𝑥𝑒 - 𝑑𝑥 = ∫ 𝑒 = 𝑑𝑢 Evaluating this integral we find:

.

∫ 𝑥𝑒 - 𝑑𝑥 =

1 = 1 . 𝑒 + 𝐶 = 𝑒- + 𝐶 2 2

We can then re-write the equation as:

.

𝑒- 𝑦 =



1 -. 𝑒 +𝐶 2

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Finally, division by the integrating factor, 𝜇(𝑥), gives “𝑦” explicitly in terms of “𝑥” and thus gives the general solution to the differential equation.

𝑦=

1 . + 𝐶𝑒 ,2 𝟏

Therefore, the correct answer choice is B. 𝐲 = + 𝐂𝐞,𝐱

𝟐

𝟐

PROBLEM 2: What is the particular solution of the following differential equation with a boundary condition 𝑦 0 = 1? 𝑑𝑦 + 𝑦 = 𝑒 ,𝑑𝑥 .

A. 𝑦 = 𝑒 ,- (𝑥 + 1) B. 𝑦 = 𝑒 - (𝑥 + 1) C. 𝑦 = 𝑒 ,- (𝑥 − 1) D. 𝑦 = 𝑒 ,- (𝑥 + 1)

SOLUTION 2: We first classify the differential equation to verify which method we should use to solve for the general and particular solutions. We see that the differential equation is a first order linear differential equation.



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The STANDARD FORM OF A FIRST ORDER LINEAR DIFFERENTIAL EQUATION is not provided in the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. We must memorize this formula and understand its application independent of the NCEES Supplied Reference Handbook. The next step in this problem is compare the differential equation with the standard form for a first order linear differential equation. 𝑑𝑦 + 𝑃 𝑥 𝑦 = 𝑄(𝑥) 𝑑𝑥 𝑑𝑦 + 𝑦 = 𝑒 ,𝑑𝑥 After writing the equation in standard form, the 𝑃(𝑥) term can be identified. We see that this equation is already expressed in standard with form the standard terms be represented as: 𝑃 𝑥 = 1 𝑎𝑛𝑑 𝑄 𝑥 = 𝑒 ,The FORMULA FOR THE INTEGRATING FACTOR is not provided in the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. We must memorize this formula and understand its application independent of the NCEES Supplied Reference Handbook. Once we have identified the 𝑃 𝑥 term, we can determine the integrating factor by plugging the 𝑃(𝑥) term into the formula for the integrating factor: 𝜇 𝑥 = 𝑒∫ 9



- :-

=𝑒

' :-

= 𝑒-

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Multiplying both sides of the differential equation by the integrating factor:

𝜇 𝑥

𝑑𝑦 + 𝑦 = 𝜇(𝑥)𝑒 ,𝑑𝑥

Plugging in the function for the integrating factor:

𝑒-

𝑑𝑦 + 𝑒 - 𝑦 = 𝑒 - 𝑒 ,𝑑𝑥

𝑒 - 𝑒 ,- = 𝑒 -,- = 𝑒 F = 1 Multiply the equation in standard form by the integrating factor, 𝜇 𝑥 and verify that the left side becomes the product rule (𝜇 𝑥 𝑦 𝑥 )′ and write it as such. The integrating factor is defined so that equation becomes equivalent to: 𝑑 𝜇 𝑥 𝑦 = 𝜇 𝑥 𝑄(𝑥) 𝑑𝑥 𝑑 𝑒 𝑦 =1 𝑑𝑥 𝑑 𝑒 - 𝑦 = 1 𝑑𝑥 We then integrate both sides of the equation with respect to the independent variable, which is “𝑥” in this differential equation. 𝜇 𝑥 𝑦 = ∫ 𝜇 𝑥 𝑄 𝑥 𝑑𝑥



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∫ 𝑑 𝑒-𝑦 =

1 𝑑𝑥

Integrating both sides gives the solution: 𝑒-𝑦 = 𝑥 + 𝐶 Finally, division by the integrating factor, 𝜇(𝑥), gives 𝑦 explicitly in terms of “𝑥”, and thus gives the general solution to the differential equation. 𝑦 = 𝑒 ,- (𝑥 + 𝐶) As we have the general solution of the differential equation, we can now use the initial condition to solve for the constant of integration, and the particular solution of the differential equation. Plugging in the initial condition of 𝑦 (0) = 1 with the values 𝑥 = 0 and 𝑦 = 1, we find: 1 = 𝑒 F (0 + 𝐶) Solving for the constant of integration we find: 𝐶=1



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We can now plug the constant of integration into our general solution and find the particular solution: 𝑦 = 𝑒 ,- (𝑥 + 1)

Therefore, the correct answer choice is D. 𝐲 = 𝐞,𝐱 (𝐱 + 𝟏) PROBLEM 3: Determine the general solution for the given differential equation and initial condition below.

𝑥

𝑑𝑦 − 2𝑦 = 𝑥 G sin 𝑥 𝑑𝑥

A. 𝑦 = −𝑥 K cos 𝑥 − 𝑥 ( sin 𝑥 + 𝐶 B. 𝑦 = −𝑥 K cos 𝑥 + 𝑥 ( sin 𝑥 + 𝐶𝑥 ( C. 𝑦 = 𝑥 K cos 𝑥 + 𝑥 ( sin 𝑥 + 𝐶𝑥 ( D. 𝑦 = −𝑥 K cos 𝑥 − 𝑥 ( sin 𝑥 + 𝐶𝑥 (

SOLUTION 3: We first classify the differential equation to verify which method we should use to solve for the general and particular solutions. We see that the differential equation is a first order linear differential equation.



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The STANDARD FORM OF A FIRST ORDER LINEAR DIFFERENTIAL EQUATION is not provided in the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. We must memorize this formula and understand its application independent of the NCEES Supplied Reference Handbook. The next step in this problem is compare the differential equation with the standard form for a first order linear differential equation. 𝑑𝑦 + 𝑃 𝑥 𝑦 = 𝑄(𝑥) 𝑑𝑥 After writing the equation in standard form, the 𝑃(𝑥) term can be identified. We rearrange the differential equation to match the standard form by dividing the both sides by “𝑥”.

𝑑𝑦 2 − 𝑦 = 𝑥 K sin 𝑥 𝑑𝑥 𝑥 Where we find the terms are defined as: 2 𝑃 𝑥 = − 𝑎𝑛𝑑 𝑄 𝑥 = 𝑥 K sin 𝑥 𝑥

The FORMULA FOR THE INTEGRATING FACTOR is not provided in the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. We must memorize this formula and understand its application independent of the NCEES Supplied Reference Handbook.



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Once we have identified the 𝑃 𝑥 term, we can determine the integrating factor by plugging the 𝑃(𝑥) term into the formula for the integrating factor:

𝜇 𝑥 = 𝑒∫ 9

- :-

= 𝑒 ,(

:-

= 𝑒 ,(NO- = 𝑒 PQ -

R.

=

1 𝑥(

Multiplying both sides of the differential equation by the integrating factor:

𝜇 𝑥

𝑑𝑦 2 − 𝑦 = 𝜇(𝑥)𝑥 K sin 𝑥 𝑑𝑥 𝑥

Plugging in the function for the integrating factor: 1 𝑑𝑦 2 1 K − 𝑦 = 𝑥 sin 𝑥 𝑥 ( 𝑑𝑥 𝑥 𝑥( Expanding out the expressions and grouping like terms we find: 1 𝑑𝑦 2 − 𝑦 = 𝑥 sin 𝑥 𝑥 ( 𝑑𝑥 𝑥 K Multiply the equation in standard form by the integrating factor, 𝜇 𝑥 and verify that the left side becomes the product rule (𝜇 𝑥 𝑦 𝑥 )′ and write it as such. The integrating factor is defined so that equation becomes equivalent to: 𝑑 𝜇 𝑥 𝑦 = 𝜇 𝑥 𝑄(𝑥) 𝑑𝑥



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𝑑 1 𝑦 = 𝑥 sin 𝑥 𝑑𝑥 𝑥 (

𝑑

1 𝑦 = 𝑥 sin 𝑥 𝑑𝑥 𝑥(

We then integrate both sides of the equation with respect to the independent variable, which is “𝑥” in this differential equation. 𝜇 𝑥 𝑦 = ∫ 𝜇 𝑥 𝑄 𝑥 𝑑𝑥

∫𝑑

1 𝑦 = ∫ x sin 𝑥 𝑑𝑥 𝑥(

The TOPIC OF INTEGRATION BY PARTS can be referenced under the topic of INTEGRAL CALCULUS on page 27 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. Using integration by parts, we evaluate the integration for the ∫ 𝜇 𝑥 𝑄 𝑥 𝑑𝑥 portion of the differential equation.

∫𝑢

𝑑𝑣 𝑑𝑢 𝑑𝑥 = 𝑢𝑣 − ∫ 𝑣 𝑑𝑥 𝑑𝑥 𝑑𝑥

Where we define “𝑢” and 𝑑𝑣/𝑑𝑥 as: 𝑢=𝑥

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𝑑𝑣 = sin 𝑥 𝑑𝑥 Integrating both sides of the differential equation with respect to “𝑥”, we find: 𝑦 = −𝑥 cos 𝑥 − ∫ 1(− cos 𝑥)𝑑𝑥 + 𝐶 𝑥( 𝑦 = −𝑥 cos 𝑥 + sin 𝑥 + 𝐶 𝑥( Finally, division by the integrating factor, 𝜇(𝑥), gives “𝑦” explicitly in terms of “𝑥” and thus gives the general solution to the differential equation. 𝑦 = −𝑥 K cos 𝑥 + 𝑥 ( sin 𝑥 + 𝐶𝑥 (

Therefore, the correct answer choice is B. 𝐲 = −𝐱 𝟑 𝐜𝐨𝐬 𝐱 + 𝐱 𝟐 𝐬𝐢𝐧 𝐱 + 𝐂𝐱 𝟐



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PROBLEM 4: What is the particular solution of the following differential equation with a boundary condition 𝑦 1 = 3?

𝑥

𝑑𝑦 + 2𝑦 = 10𝑥 ( 𝑑𝑥

A. 𝑦 =

' (

B. 𝑦 = − C. 𝑦 =

5𝑥 ( +

' G

'

-.

5𝑥 ( +

(

5𝑥 ( +

D. . 𝑦 = −

'

' G

' -.

' -.

5𝑥 ( +

' -.

SOLUTION 4: We first classify the differential equation to verify which method we should use to solve for the general and particular solutions. We see that the differential equation is a first order linear differential equation. The STANDARD FORM OF A FIRST ORDER LINEAR DIFFERENTIAL EQUATION is not provided in the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. We must memorize this formula and understand its application independent of the NCEES Supplied Reference Handbook.



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The next step in this problem is compare the differential equation with the standard form for a first order linear differential equation. 𝑑𝑦 + 𝑃 𝑥 𝑦 = 𝑄(𝑥) 𝑑𝑥 𝑥

𝑑𝑦 + 2𝑦 = 10𝑥 ( 𝑑𝑥

After writing the equation in standard form, the 𝑃(𝑥) term can be identified. We see rearrange the equation to be in the standard with form the standard terms be represented as: 𝑑𝑦 2 + 𝑦 = 10𝑥 𝑑𝑥 𝑥 𝑃 𝑥 =

2 𝑎𝑛𝑑 𝑄 𝑥 = 10𝑥 𝑥

Once we have identified the 𝑃 𝑥 term, we can determine the integrating factor by plugging the 𝑃(𝑥) term into the formula for the integrating factor:

𝜇 𝑥 = 𝑒∫ 9

- :-

=𝑒

( :-

.

= 𝑒 ( PQ - = 𝑒 PQ - = 𝑥 (

Multiplying both sides of the differential equation by the integrating factor:

𝜇 𝑥



𝑑𝑦 2 + 𝑦 = 𝜇(𝑥)10𝑥 𝑑𝑥 𝑥

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Plugging in the function for the integrating factor:

𝑥(

𝑑𝑦 2 + 𝑦 = 𝑥 ( 10𝑥 𝑑𝑥 𝑥

We simplify the expression and combine like terms to find:

𝑥(

𝑑𝑦 + 2𝑥𝑦 = 10𝑥 K 𝑑𝑥

Multiply the equation in standard form by the integrating factor, 𝜇 𝑥 and verify that the left side becomes the product rule (𝜇 𝑥 𝑦 𝑥 )′ and write it as such. The integrating factor is defined so that equation becomes equivalent to: 𝑑 𝜇 𝑥 𝑦 = 𝜇 𝑥 𝑄(𝑥) 𝑑𝑥 𝑑 ( 𝑥 𝑦 = 10𝑥 K 𝑑𝑥 𝑑 𝑥 ( 𝑦 = 10𝑥 K 𝑑𝑥 We then integrate both sides of the equation with respect to the independent variable, which is “𝑥” in this differential equation. 𝜇 𝑥 𝑦 = ∫ 𝜇 𝑥 𝑄 𝑥 𝑑𝑥

∫ 𝑑 𝑥(𝑦 =

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Integrating both sides gives the solution: 𝑥G 𝑥 𝑦 = 10 + 𝐶 4 (

Finally, division by the integrating factor, 𝜇(𝑥), gives “𝑦” explicitly in terms of “𝑥”, and thus gives the general solution to the differential equation.

𝑦=

5 ( 𝐶 𝑥 + ( 2 𝑥

As we have the general solution of the differential equation, we can now use the initial condition to solve for the constant of integration, and the particular solution of the differential equation. Plugging in the initial condition of 𝑦 (1) = 3 with the values 𝑥 = 1 and 𝑦 = 3, we find:

3=

5 𝐶 1+ 2 1

Solving for the constant of integration we find:

𝐶=



1 2

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We can now plug the constant of integration into our general solution and find the particular solution: 1 5 ( 1 1 𝑦 = 𝑥 + 2( = 5𝑥 ( + ( 2 𝑥 2 𝑥

Therefore, the correct answer choice is A. 𝐲 =



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𝟏 𝐱𝟐