- #1
Lancelot59
- 646
- 1
I need to find a solution to the following problem:
[tex](x^{2}-1)\frac{dy}{dx}+2y=(x+1)^{2}[/tex]
[tex]y(0)=0[/tex]
I decided to try using variation of parameters. My teacher was unable to show any examples, and I'm having issues understanding the textbook.
From what I see I need to get it onto this form:
[tex]y'=f(x)y+g(x)[/tex]
I think this is correct
[tex]\frac{dy}{dx}=\frac{(x+1)^{2}-2y}{x^{2}-1}=\frac{-2}{x^{2}-1}y+\frac{(x+1)^{2}}{x^{2}-1}[/tex]
Now what do I need to do? I'm having trouble understanding the textbook.
[tex](x^{2}-1)\frac{dy}{dx}+2y=(x+1)^{2}[/tex]
[tex]y(0)=0[/tex]
I decided to try using variation of parameters. My teacher was unable to show any examples, and I'm having issues understanding the textbook.
From what I see I need to get it onto this form:
[tex]y'=f(x)y+g(x)[/tex]
I think this is correct
[tex]\frac{dy}{dx}=\frac{(x+1)^{2}-2y}{x^{2}-1}=\frac{-2}{x^{2}-1}y+\frac{(x+1)^{2}}{x^{2}-1}[/tex]
Now what do I need to do? I'm having trouble understanding the textbook.