69.1 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: The general solution of the following differential equation is best represented as: 𝑦 ! + 2𝑥𝑦 = 𝑥 !

A. 𝑦(𝑥) = − 𝐶𝑒 !!

!

! !

B. 𝑦(𝑥) = + 𝐶𝑒 !!

!

! !

C. 𝑦(𝑥) = + 𝐶𝑒 !

!

! !

D. 𝑦(𝑥) = ! − 𝐶𝑒 !!



!

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PROBLEM 2: The particular solution of the following differential equation, with a boundary condition 𝑦 0 = 1, is best represented as: 𝑑𝑦 + 𝑦 = 𝑒 !! 𝑑𝑥 !

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



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PROBLEM 3: The general solution for the given differential equation is best represented as:

𝑥

𝑑𝑦 − 2𝑦 = 𝑥 ! sin 𝑥 𝑑𝑥

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



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PROBLEM 4: The particular solution of the following differential equation with a boundary condition 𝑦 1 = 3 is most close to:

𝑥

𝑑𝑦 + 2𝑦 = 10𝑥 ! 𝑑𝑥 !

!

A. 𝑦 = ! 5𝑥 ! + ! ! !

!

B. 𝑦 = − ! 5𝑥 ! + ! ! !

!

C. 𝑦 = ! 5𝑥 ! + ! ! D. . 𝑦 = −



! !

5𝑥 ! +

! !!

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FIRST ORDER LINEAR DIFFERENTIAL EQUATIONS | SOLUTIONS SOLUTION 1: The first thing we need to do here is classify the differential equation to verify which method we should use to solve for the general and particular solutions. In this problem, we are given the function: 𝑦 ! + 2𝑥𝑦 = 𝑥 This DIFFERENTIAL EQUATION is both FIRST ORDER and LINEAR. 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. A FIRST ORDER LINEAR DIFFERENTIAL EQUATION is typically arranged in the standard form: 𝑑𝑦 + 𝑃 𝑥 𝑦 = 𝑄(𝑥) 𝑑𝑥



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Where: • P x and Q x are functions of "x", and in some cases may be constants • “𝑥” is the INDEPENDENT VARIABLE • “𝑦” is the DEPENDENT VARIABLE So at this point, let’s confirm that we have it all set up in this form so we can move forward without hitting any obstacles. Using different, yet equivalent nomenclature, we can rewrite this equation to read as: 𝑑𝑦 + 2𝑥𝑦 = 𝑥 𝑑𝑥 This equation is expressed in standard form with the standard terms represented as: 𝑃 𝑥 = 2𝑥 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.



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With the 𝑃 𝑥 term now identified, we can determine the integrating factor by plugging this term into the standard form of the INTEGRATING FACTOR, which is: 𝜇 𝑥 = 𝑒∫ !

! !"

Plugging in 𝑃 𝑥 we get: 𝜇 𝑥 =𝑒

!! !"

Carrying out the INTEGRATION, our INTEGRATING FACTOR becomes: 𝜇 𝑥 = 𝑒!

!

We now 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. 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 and expanding our expression, we get: !

!

𝑒 ! 𝑦 ! + 2𝑥𝑒 ! 𝑦 = 𝑥𝑒 !



!

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We need to verify that the left side becomes the product rule (𝜇 𝑥 𝑦 𝑥 )′ and write it as such. Viewing our formula at this point, we can confirm that indeed the left side follows along with the PRODUCT RULE, which again reads as: The First, times the Derivative of the Second, plus the second times the derivative of the First Now adjusting the formula so that the left side is written in the form (𝜇 𝑥 𝑦 𝑥 )′ we get: 𝑑 !! ! 𝑒 𝑦′ = 𝑥𝑒 ! 𝑑𝑥 We now need to INTEGRATE both sides of the equation with respect to the independent variable, which is “𝑥” in this differential equation. Rearranging a bit first, we get: !

!

𝑑 𝑒 ! 𝑦′ = 𝑥𝑒 ! 𝑑𝑥 And INTEGRATING: !

!

∫ 𝑑 𝑒 ! 𝑦′ = ∫ 𝑥𝑒 ! 𝑑𝑥



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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. The left side gives us no heart ache, but the right side of this INTEGRATION may present some difficulty if we didn’t know what we were doing. However, we will use INTEGRATION BY PARTS, to evaluate the right side: !

∫ 𝑥𝑒 ! 𝑑𝑥 INTEGRATION BY PARTS lays out as:

∫𝑢

𝑑𝑣 𝑑𝑢 𝑑𝑥 = 𝑢𝑣 − ∫ 𝑣 𝑑𝑥 𝑑𝑥 𝑑𝑥

In our case, we will define: 𝑢 = 𝑥! And: 𝑑𝑢 = 2𝑥 𝑑𝑥



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We plug in these values: !

∫ 𝑥𝑒 ! 𝑑𝑥 = ∫ 𝑒 ! 𝑑𝑢 And evaluate the integral:

∫ 𝑒 ! 𝑑𝑢 =

1 ! 𝑒 +𝐶 2

Subbing back in our SUBSTITUTED values, we have: !

∫ 𝑥𝑒 ! 𝑑𝑥 =

1 !! 𝑒 +𝐶 2

Which gives us an expression after carrying out the INTEGRATION on both sides as:

!

𝑒! 𝑦 =

1 !! 𝑒 +𝐶 2

This is the IMPLICIT SOLUTION to the FIRST ORDER LINEAR DIFFERENTIAL EQUATION. We can now determine the EXPLICIT SOLUTION. Recall that the EXPLICIT SOLUTION organizes the expression so that the INDEPENDENT and DEPENDENT VARIABLES are isolated on either side of the equation and takes the strict form of 𝑦 = 𝑓(𝑥), where 𝑦 is explicitly defined by a function 𝑓 𝑥 .



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We are close here, but do not have an expression that is EXPLICIT due to the exponential function being on the left side of the equation…we have to tackle that. Doing so, we can divide by the integrating factor, 𝜇(𝑥), getting “𝑦” explicitly in terms of “𝑥", and thus, giving us the general solution to the differential equation as: !

𝑦(𝑥) = + 𝐶𝑒 !!

!

!

𝟏

The correct answer choice is B. 𝐲(𝐱) = 𝟐 + 𝐂𝐞!𝐱

𝟐

SOLUTION 2: The first thing we need to do here is classify the differential equation to verify which method we should use to solve for the general and particular solutions. In this problem, we are given the function: 𝑑𝑦 + 𝑦 = 𝑒 !! 𝑑𝑥 This DIFFERENTIAL EQUATION is both FIRST ORDER and LINEAR. 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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A FIRST ORDER LINEAR DIFFERENTIAL EQUATION is typically arranged in the standard form: 𝑑𝑦 + 𝑃 𝑥 𝑦 = 𝑄(𝑥) 𝑑𝑥 Where: • P x and Q x are functions of "x", and in some cases may be constants • “𝑥” is the INDEPENDENT VARIABLE • “𝑦” is the DEPENDENT VARIABLE So at this point, let’s confirm that we have it all set up in this form so we can move forward without hitting any obstacles. We see that this equation is already expressed in standard form with the standard terms represented as: 𝑃 𝑥 =1 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.



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With the 𝑃 𝑥 term now identified, we can determine the integrating factor by plugging this term into the standard form of the INTEGRATING FACTOR, which is: 𝜇 𝑥 = 𝑒∫ !

! !"

Plugging in 𝑃 𝑥 we get: 𝜇 𝑥 =𝑒

! !"

Carrying out the INTEGRATION, our INTEGRATING FACTOR becomes: 𝜇 𝑥 = 𝑒! We now 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. 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

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Honing in real quick on the left side of the formula, we have: 𝑒 ! 𝑒 !! By law, this can be simplified to: 𝑒 !!! And further: 𝑒! We know that the exponential function raised to the zero is: 𝑒! = 1 Therefore, our formula simplifies to:

𝑒!

𝑑𝑦 + 𝑒!𝑦 = 1 𝑑𝑥

We need to verify that the left side becomes the product rule (𝜇 𝑥 𝑦 𝑥 )′ and write it as such.



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Viewing our formula at this point, we can confirm that indeed the left side follows along with the PRODUCT RULE, which again reads as: The First, times the Derivative of the Second, plus the second times the derivative of the First Now adjusting the formula so that the left side is written in the form (𝜇 𝑥 𝑦 𝑥 )′ we get: 𝑑 ! 𝑒 𝑦 =1 𝑑𝑥 We now need to INTEGRATE both sides of the equation with respect to the independent variable, which is “𝑥” in this differential equation. Rearranging a bit first, we get: 𝑑 𝑒 ! 𝑦 = 1 𝑑𝑥 And INTEGRATING:

∫ 𝑑 𝑒!𝑦 =

1 𝑑𝑥

Evaluating each integral: 𝑒!𝑦 = 𝑥 + 𝐶

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This is the IMPLICIT SOLUTION to the FIRST ORDER LINEAR DIFFERENTIAL EQUATION. We can now determine the EXPLICIT SOLUTION. Recall that the EXPLICIT SOLUTION organizes the expression so that the INDEPENDENT and DEPENDENT VARIABLES are isolated on either side of the equation and takes the strict form of 𝑦 = 𝑓(𝑥), where 𝑦 is explicitly defined by a function 𝑓 𝑥 . We are close here, but do not have an expression that is EXPLICIT due to the exponential function being on the left side of the equation…we have to tackle that. Doing so, we can divide by the integrating factor, 𝜇(𝑥), getting “𝑦” explicitly in terms of “𝑥", and thus, giving us the general solution to the differential equation as: 𝑦(𝑥) = 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. Again, we were given the INITIAL CONDITION: 𝑦(0) = 1



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Plugging this INITIAL CONDITION in to our GENERAL SOLUTION, we have: 1 = 𝑒 ! (0 + 𝐶) Rearranging and solving for the CONSTANT OF INTEGRATION we get: 𝐶=1 We can now plug the CONSTANT OF INTEGRATION into our general solution and find the PARTICULAR SOLUTION, which is: 𝑦(𝑥) = 𝑒 !! (𝑥 + 1) The correct answer choice is D. 𝐲(𝐱) = 𝐞!𝐱 (𝐱 + 𝟏)

SOLUTION 3: The first thing we need to do here is classify the differential equation to verify which method we should use to solve for the general and particular solutions. In this problem, we are given the function:

𝑥

𝑑𝑦 − 2𝑦 = 𝑥 ! sin 𝑥 𝑑𝑥

This DIFFERENTIAL EQUATION is both FIRST ORDER and LINEAR.



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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. A FIRST ORDER LINEAR DIFFERENTIAL EQUATION is typically arranged in the standard form: 𝑑𝑦 + 𝑃 𝑥 𝑦 = 𝑄(𝑥) 𝑑𝑥 Where: • P x and Q x are functions of "x", and in some cases may be constants • “𝑥” is the INDEPENDENT VARIABLE • “𝑦” is the DEPENDENT VARIABLE So at this point, let’s confirm that we have it all set up in this form so we can move forward without hitting any obstacles. We need to get the front end differential to have a coefficient of 1, so let’s divide through by the variable x. Doing so we get: 𝑑𝑦 2 − 𝑦 = 𝑥 ! sin 𝑥 𝑑𝑥 𝑥

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This equation is expressed in standard form with the standard terms represented as:

𝑃 𝑥 =−

2 𝑥

𝑄 𝑥 = 𝑥 ! sin 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. With the 𝑃 𝑥 term now identified, we can determine the integrating factor by plugging this term into the standard form of the INTEGRATING FACTOR, which is: 𝜇 𝑥 = 𝑒∫ !

! !"

Plugging in 𝑃 𝑥 we get:

𝜇 𝑥 = 𝑒 !!

!" !

Carrying out the INTEGRATION, our INTEGRATING FACTOR becomes: 𝜇 𝑥 = 𝑒 !!!"# Which can be rewritten as: 𝜇 𝑥 = 𝑒 !" !

!!

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And simplifies to:

𝜇 𝑥 =

1 𝑥!

We now 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. Multiplying both sides of the differential equation by the integrating factor:

𝜇 𝑥

𝑑𝑦 2 − 𝑦 = 𝜇(𝑥)𝑥 ! sin 𝑥 𝑑𝑥 𝑥

Plugging in the function for the integrating factor: 1 𝑑𝑦 2 1 ! − 𝑦 = 𝑥 sin 𝑥 𝑥 ! 𝑑𝑥 𝑥 𝑥! Expanding and simplifying our expression, we get: 1 𝑑𝑦 2 − 𝑦 = 𝑥 sin 𝑥 𝑥 ! 𝑑𝑥 𝑥 ! We need to verify that the left side becomes the product rule (𝜇 𝑥 𝑦 𝑥 )′ and write it as such. Viewing our formula at this point, we can confirm that indeed the left side follows along with the PRODUCT RULE, which again reads as:

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The First, times the Derivative of the Second, plus the second times the derivative of the First Now adjusting the formula so that the left side is written in the form (𝜇 𝑥 𝑦 𝑥 )′ we get: 𝑑 1 𝑦 = 𝑥 sin 𝑥 𝑑𝑥 𝑥 ! We now need to INTEGRATE both sides of the equation with respect to the independent variable, which is “𝑥” in this differential equation. Rearranging a bit first, we get:

𝑑

1 𝑦 = 𝑥 sin 𝑥 𝑑𝑥 𝑥!

And INTEGRATING:

∫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. The left side gives us no heart ache, but the right side of this INTEGRATION may present some difficulty if we didn’t know what we were doing.

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However, we will use INTEGRATION BY PARTS, to evaluate the right side: ∫ x sin 𝑥 𝑑𝑥 INTEGRATION BY PARTS lays out as:

∫𝑢

𝑑𝑣 𝑑𝑢 𝑑𝑥 = 𝑢𝑣 − ∫ 𝑣 𝑑𝑥 𝑑𝑥 𝑑𝑥

In our case, we will define: 𝑢=𝑥 And: 𝑑𝑣 = sin 𝑥 𝑑𝑥 We plug in these values: ∫ x sin 𝑥 𝑑𝑥 = −𝑥 cos 𝑥 − ∫ 1(− cos 𝑥)𝑑𝑥 + 𝐶 And evaluate the integral: ∫ x sin 𝑥 𝑑𝑥 = −𝑥 cos 𝑥 + sin 𝑥 + 𝐶



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Which gives us an expression after carrying out the INTEGRATION on both sides as: 𝑦 = −𝑥 cos 𝑥 + sin 𝑥 + 𝐶 𝑥! This is the IMPLICIT SOLUTION to the FIRST ORDER LINEAR DIFFERENTIAL EQUATION. We can now determine the EXPLICIT SOLUTION. Recall that the EXPLICIT SOLUTION organizes the expression so that the INDEPENDENT and DEPENDENT VARIABLES are isolated on either side of the equation and takes the strict form of 𝑦 = 𝑓(𝑥), where 𝑦 is explicitly defined by a function 𝑓 𝑥 . We are close here, but do not have an expression that is EXPLICIT due to the x squared term being on the left side of the equation…we have to tackle that. Doing so, we can multiply both sides by this term, getting “𝑦” explicitly in terms of “𝑥", and thus, giving us the general solution to the differential equation as: 𝑦(𝑥) = −𝑥 ! cos 𝑥 + 𝑥 ! sin 𝑥 + 𝐶𝑥 ! The correct answer choice is B. 𝐲(𝐱) = −𝐱 𝟑 𝐜𝐨𝐬 𝐱 + 𝐱 𝟐 𝐬𝐢𝐧 𝐱 + 𝐂𝐱 𝟐



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SOLUTION 4: The first thing we need to do here is classify the differential equation to verify which method we should use to solve for the general and particular solutions. In this problem, we are given the function:

𝑥

𝑑𝑦 + 2𝑦 = 10𝑥 ! 𝑑𝑥

This DIFFERENTIAL EQUATION is both FIRST ORDER and LINEAR. 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. A FIRST ORDER LINEAR DIFFERENTIAL EQUATION is typically arranged in the standard form: 𝑑𝑦 + 𝑃 𝑥 𝑦 = 𝑄(𝑥) 𝑑𝑥 Where: • P x and Q x are functions of "x", and in some cases may be constants • “𝑥” is the INDEPENDENT VARIABLE

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• “𝑦” is the DEPENDENT VARIABLE So at this point, let’s confirm that we have it all set up in this form so we can move forward without hitting any obstacles. We need to get the front end differential to have a coefficient of 1, so let’s divide through by the variable x. Doing so we get: 𝑑𝑦 2 + 𝑦 = 10𝑥 𝑑𝑥 𝑥 This equation is expressed in standard form with the standard terms represented as:

𝑃 𝑥 =

2 𝑥

𝑄 𝑥 = 10𝑥 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. With the 𝑃 𝑥 term now identified, we can determine the integrating factor by plugging this term into the standard form of the INTEGRATING FACTOR, which is: 𝜇 𝑥 = 𝑒∫ !

! !"

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Plugging in 𝑃 𝑥 we get:

𝜇 𝑥 =𝑒

! !" !

Carrying out the INTEGRATION, our INTEGRATING FACTOR becomes: 𝜇 𝑥 = 𝑒 ! !" ! Which can be rewritten as: 𝜇 𝑥 = 𝑒 !" !

!

And further simplified as: 𝜇 𝑥 = 𝑥! We now 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. 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

And multiplying everything out, we get:

𝑥!

𝑑𝑦 + 2𝑥𝑦 = 10𝑥 ! 𝑑𝑥

We need to verify that the left side becomes the product rule (𝜇 𝑥 𝑦 𝑥 )′ and write it as such. Viewing our formula at this point, we can confirm that indeed the left side follows along with the PRODUCT RULE, which again reads as: The First, times the Derivative of the Second, plus the second times the derivative of the First Now adjusting the formula so that the left side is written in the form (𝜇 𝑥 𝑦 𝑥 )′ we get: 𝑑 ! 𝑥 𝑦 = 10𝑥 ! 𝑑𝑥 We now need to INTEGRATE both sides of the equation with respect to the independent variable, which is “𝑥” in this differential equation.



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Rearranging a bit first, we get: 𝑑 𝑥 ! 𝑦 = 10𝑥 ! 𝑑𝑥 And INTEGRATING:

∫ 𝑑 𝑥!𝑦 =

10𝑥 ! 𝑑𝑥

Evaluating each integral: 𝑥! 𝑥 𝑦 = 10 + 𝐶 4 !

This is the IMPLICIT SOLUTION to the FIRST ORDER LINEAR DIFFERENTIAL EQUATION. We can now determine the EXPLICIT SOLUTION. Recall that the EXPLICIT SOLUTION organizes the expression so that the INDEPENDENT and DEPENDENT VARIABLES are isolated on either side of the equation and takes the strict form of 𝑦 = 𝑓(𝑥), where 𝑦 is explicitly defined by a function 𝑓 𝑥 . We are close here, but do not have an expression that is EXPLICIT due to the x squared term being on the left side of the equation…we have to tackle that.



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Doing so, we can divide by x squared getting “𝑦” explicitly in terms of “𝑥", and thus, giving us the general solution to the differential equation as:

𝑦(𝑥) =

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. Again, we were given the INITIAL CONDITION: 𝑦(1) = 3 Plugging this INITIAL CONDITION in to our GENERAL SOLUTION, we have:

3=

5 𝐶 1+ 2 1

Rearranging and solving for the CONSTANT OF INTEGRATION we get:

𝐶=



1 2

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We can now plug the CONSTANT OF INTEGRATION into our general solution and find the PARTICULAR SOLUTION, which is: 1 5 ! 𝑦(𝑥) = 𝑥 + 2! 2 𝑥 Which simplifies to:

𝑦(𝑥) =

1 1 5𝑥 ! + ! 2 𝑥

The correct answer choice is A. 𝐲 =



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𝟏 𝟐

𝟓𝐱 𝟐 +

𝟏 𝐱𝟐

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