Expert: Paul Klarreich - 7/5/2009

Question
How can i find the general solution to the First Order DE as dy/dx = 2y + x[(e^(3x) - e^(2x)].

I used integrating factor e^(integral -2 dx) = e^(-2x).

Multiply both sides by e^(-2x):
e^(-2x)dy/dx - 2e^(-2x)y = xe^x - x.
==> (d/dx) [e^(-2x) y] = xe^x - x.

Integrate both sides (and using integration by parts on the right side):
e^(-2x) y = (xe^x - e^x) - x^2/2 + C.

Therefore,
y = e^(3x)(x - 1) + x^2 e^(2x)/2 + Ce^(2x).

Use y(0) = 2 to find C:
2 = -1 + 0 + C ==> C = 3.

So, my solution is y = e^(3x)(x - 1) + x^2 e^(2x)/2 + 3e^(2x).

But when i try to differentiate it it is not right i think. Pls tell me where did i go wrong.

Thanks.

Answer
Questioner:   jim
Country:  United States
Category:  Advanced Math
Private:  No
 
Subject:  calculus
Question:  How can i find the general solution to the First Order DE as dy/dx = 2y + x[(e^(3x) - e^(2x)].

I used integrating factor e^(integral -2 dx) = e^(-2x).

Multiply both sides by e^(-2x):
e^(-2x)dy/dx - 2e^(-2x)y = xe^x - x.
==> (d/dx) [e^(-2x) y] = xe^x - x.

Integrate both sides (and using integration by parts on the right side):
e^(-2x) y = (xe^x - e^x) - x^2/2 + C.

Therefore,
y = e^3x (x - 1) + x^2 e^(2x)/2 + Ce^(2x).

Use y(0) = 2 to find C:
2 = -1 + 0 + C ==> C = 3.

So, my solution is y = e^(3x)(x - 1) + x^2 e^(2x)/2 + 3e^(2x).

But when i try to differentiate it it is not right i think. Pls tell me where did i go wrong.

Thanks.
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dy/dx = 2y + x[(e^(3x) - e^(2x)]

e^(integral -2 dx) = e^(-2x)

dy/dx e^(-2x) = 2y e^(-2x) + x[(e^(3x)e^(-2x) - e^(2x)e^(-2x)]

dy/dx e^(-2x) = 2y e^(-2x) + x[(e^x - 1]

dy/dx e^(-2x) - 2y e^(-2x) + x[(e^x - 1]

D(e^(-2x) y) = x[e^x - 1]

e^(-2x) y = INT (x e^x - x)

e^(-2x) y = x e^x - e^x - x^2/2 + C.

y = e^3x(x - 1) - e^2x x^2/2 + C e^2x

>> Yes, yes, with yours my figures do agree.
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Check, please?  [sorry -- I just got back from dinner out.]

y' = 3e^3x(x - 1) + e^3x - e^2x x^2 - e^2x x + 2C e^2x

y' = 3e^3x x - 3e^3x + e^3x - e^2x x^2 - e^2x x + 2C e^2x

Now  y' - 2y =

3e^3x x - 3e^3x + e^3x - e^2x x^2 - e^2x x + 2C e^2x
- 2(e^3x(x - 1) - e^2x x^2/2 + C e^2x)

3e^3x x - 3e^3x + e^3x - e^2x x^2 - e^2x x + 2C e^2x
 - 2 x e^3x + 2e^3x + e^2x x^2 + 2C e^2x

e^3x x - e^2x x

x(e^3x - e^2x)

Yes, yes, it checks.  So what could be wrong?

Use y(0) = 2 to find C:

2 = 1(0 - 1) - 1 (0)/2 + C (1)

2 = -1 + C

C = 3

Sorry, but it all looks OK to me.

Maybe your check didn't work out correctly.  But if it checks for any C, then it surely would check for  C = 3.
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A last idea -- maybe you didn't get the answer in the back of the book?  Remember that the answer might be the same as yours but in a different form.