Question

Need some help please..
Use Variation of Parameters:
y''-y=e^x+1
I need some help getting started with Yp please

Answer 1

The problem shows as:
\[y''-y=e^x+1\]
I havent done these before so I would like some help for the first couple problems..

Answer 2

Eh, why do we have to use variation of parameters?

Answer 3

Its the technique im trying to learn..

Answer 4

there is another problem:
\[y''+y=cscx\]

Answer 5

You have a differential equation of the form:
y'' + q(x)y' + r(x)y = g(x)
First, identify q(x) and r(x), and g(x).

Answer 6

How do I do that?

Answer 7

Would you mind helping me out for a sec? @AccessDenied

Answer 8

If you're supposed to use variation of parameters, you have to find two linearly independent solutions. You'll get those from the characteristic solution.

Answer 9

@jon91d, you do know how to find the general characteristic solution, don't you?

Answer 10

You mean the Yc?
\[Yc=C1e^{-x}+C2e^x\]

Answer 11

Yes. The set up for VoP requires taking two known solutions, call them \(y_1\) and \(y_2\). In this case, \(y_1=e^{-x}\) and \(y_2=e^x\).

Answer 12

Ok, gotcha, is this where you need to apply the Wronskian?

Answer 13

Almost there. The solution to the equation will be of the form,
\[y=u_1y_y+u_2y_2\]
The formulas for the \(u\)'s involves taking the Wronskian.
\[u_1=-\int\frac{y_2f(x)}{W(y_1,y_2)}~dx\\
u_2=-\int\frac{y_1f(x)}{W(y_1,y_2)}~dx\]
where \(f(x)\) comes from the nonhomogeneous part of the original equation. So that would be \(f(x)=e^x+1\). The Wronskian would be the determinant,
\[W(y_1,y_2)=\begin{vmatrix}y_1&y_2\\\frac{dy_1}{dx}&\frac{dy_2}{dx}\end{vmatrix}\]

Answer 14

So the Wronskian would be?
d