Solving a first order DE

• Sep 6th 2009, 07:33 AM
sgupta
Solving a first order DE
Hello,

I am trying to solve this problem:
x(dy/dx)-2y=x^15
By bringing this to the standard form I get:
(dy/dx)-(2/x)y=x^14
Now I multiply the equation by (1/x^2)
I get the form d/dx(y/x^2)=(x^12)
Integrating both sides I get: (y/x^2)=((x^13)/13)+C(x^2)

Now I tried using the condition y(1)=-2 to get the value of C. However, every time I get the value of C to be 1, unfortunately this is not the right answer.

Where am I going wrong, any help would be highly appreciated.

Thanks!
• Sep 6th 2009, 08:15 AM
CaptainBlack
Quote:

Originally Posted by sgupta
Hello,

I am trying to solve this problem:
x(dy/dx)-2y=x^15
By bringing this to the standard form I get:
(dy/dx)-(2/x)y=x^14
Now I multiply the equation by (1/x^2)
I get the form d/dx(y/x^2)=(x^12)
Integrating both sides I get: (y/x^2)=((x^13)/13)+C(x^2)

Now I tried using the condition y(1)=-2 to get the value of C. However, every time I get the value of C to be 1, unfortunately this is not the right answer.

Where am I going wrong, any help would be highly appreciated.

Thanks!

Assuming what you have for the general solution is correct:

$\frac{y}{x^2}=\frac{x^{13}}{13}+Cx^2$

Put $x=1$:

$y(1)=-2=\frac{1}{13}+C$

CB