Y'' + y = f(x) - Variation of Parameters?

[FeDeX_LaTeX]

y'' + y = f(x) -- Variation of Parameters?

Homework Statement

Use variation of parameters to solve

$y'' + y = f(x), y(0) = y'(0) = 0.$

Homework Equations

A description of the method is here: http://en.wikipedia.org/wiki/Variation_of_parameters

The Attempt at a Solution

The complementary function is:

$y_{CF} = A\cos(x) + B\sin(x) = Ay_1 + By_2$, say.

If $y = A(x)y_{1}(x) + B(x)y_{2}(x)$ is the solution of the forced equation, then, if we choose $A'y_1 + B'y_2 = 0$, $y' = Ay'_{1} + By'_{2}$ and $y'' = A'y'_{1} + B'y'_{2}$.

Since $y_{1,2}$ satisfy the homogeneous equation, substitution will lead to $A'y'_{1} + B'y'_{2} = f(x)$.

Solving these two equations for A' and B' yields

$A'(y_{2}y'_{1} - y_{1}y'_{2}) = f(x)y_{2}$ and
$B'(y_{2}y'_{1} - y_{1}y'_{2}) = -f(x)y_{2}$

Now, if $y_{1} = \cos(x), y_{2} = \sin(x)$ then differentiating and substituting gives us

$A' = -f(x) \implies A = \int -f(x) dx$ and
$B' = f(x) \implies B = \int f(x) dx$.

But from here, I'm stuck. Any advice?

EDIT: Am I going about this the wrong way? I'm following the example at the top of page 3 here.

 

[FeDeX_LaTeX]

Anyone? (Sorry, this homework is due pretty soon...)

 

[vanhees71]

Well, just solve your linear equations for $A'$ and $B'$ and then integrate. You are pretty close to the solution!

 

[FeDeX_LaTeX]

I've ended up with the particular integral:

$y = (\sin x - \cos x)\int f(x) dx$

But I think this is incorrect -- WolframAlpha is having sines and cosines in the integral itself. Where are those coming from?

 

[ehild]

Homework Statement

Use variation of parameters to solve

$y'' + y = f(x), y(0) = y'(0) = 0.$

The Attempt at a Solution

The complementary function is:

$y_{CF} = A\cos(x) + B\sin(x) = Ay_1 + By_2$, say.

If $y = A(x)y_{1}(x) + B(x)y_{2}(x)$ is the solution of the forced equation, then, if we choose $A'y_1 + B'y_2 = 0$, $y' = Ay'_{1} + By'_{2}$ and $y'' = A'y'_{1} + B'y'_{2}$.

Correctly: $y'' = A'y'_{1} + B'y'_{2}+Ay''_{1}+By''_{2}$

Since $y_{1,2}$ satisfy the homogeneous equation, substitution will lead to $A'y'_{1} + B'y'_{2} = f(x)$.

Solving these two equations for A' and B' yields

$A'(y_{2}y'_{1} - y_{1}y'_{2}) = f(x)y_{2}$ and
$B'(y_{2}y'_{1} - y_{1}y'_{2}) = -f(x)y_{2}$

Now, if $y_{1} = \cos(x), y_{2} = \sin(x)$ then differentiating and substituting gives us

$A' = -f(x) \implies A = \int -f(x) dx$ and
$B' = f(x) \implies B = \int f(x) dx$.

.

It is $$A' = \frac{f(x)y_2}{y_1'y_2-y_1y_2'}=-f(x)\sin(x)$$ and $$B' = -\frac{f(x)y_1}{y_1'y_2-y_1y_2'}=f(x)\cos(x)$$ehild