Y'+8y=2xe^(-5x); y(0)=C [Differential Equations Problem]

When I first tried to solve this, I used the following theorem:
y'+P(x)y=Q(x), and

1/(e^(integral of P(x))*(integral of [Q(x)*(e^(integral of P(x)]).

I did so because I could not find a way to separate the variables. Thus I defined 8 as P(x), which gave me e^(8x) after integration. Therefore Q(x) was defined as 2xe^(-5x), which was then multiplied by e^(8x).

In solving this I got (2/9)*(e^(-5x))*(3x-1)

y' + 8y = 2x*e^(-5x), y(0) = C

Integrating factor = e^(∫8 dx) = e^(8x)

Multiplying both sides by integrating factor:

y'(e^(8x)) + 8e^(8x)y = 2x*e^(3x)

d/dx[y*e^(8x)] = 2x*e^(3x)

Integrating both sides:

y*e^(8x) = ∫2x*e^(3x) dx

Integrating by parts:

u = 2x => du = 2 dx

dv = e^(3x) dx => v = e^(3x)/3

y*e^(8x) = 2x*e^(3x)/3 - 2/3*∫e^(3x) dx

y*e^(8x) = 2x*e^(3x)/3 - 2/9*e^(3x) + D

y = e^(-5x)*(2x/3 - 2/9 + D/e^(3x))

C = D - 2/9 => D = C + 2/9

y = e^(-5x)*(2x/3 - 2/9 + (C + 2/9)*e^(-3x))

Why thank you George.. I'm glad this helped.

Report Abuse