# Thread: Superposition principle-homogeneous equation

1. ## Superposition principle-homogeneous equation

1. Show that y1=cosx and y2=sinx are solution of the DE y''+y=0.verify that ⱷ(x)=c1cosx+c2sinx is also a solution of the DE.
2. Given that y1 and y2 are the solution of the DE y''+y^2=0. Is y3 =c1y1+c2y2 also a solution of the DE?

2. ## Re: Superposition principle-homogeneous equation

Just to let the helpers here know, this question has also been posted at MMF, and I have offered hints on how to proceed.

For the first one, you will need to find $\varphi''(x)$ and substitute into the given ODE.

For the second one, you are told:

$y_1''+y_1^2=0$

$y_2''+y_2^2=0$

Multiply the first equation by $c_1$ and the second by $c_2$, then add the equations. What do you find?

wat is mmf?

4. ## Re: Superposition principle-homogeneous equation

Originally Posted by cherrytreefriend
wat is mmf?
It is one of the other forums on which you have posted this question, mymathforum.com.

We try to look out for each other at the different sites when we notice questions posted by the same user at multiple sites to prevent duplication of effort.

5. ## Re: Superposition principle-homogeneous equation

for the 2nd...i got until c1y1''+c1y1^2+c2y2"+c2y2^2 after adding two equation..

6. ## Re: Superposition principle-homogeneous equation

What does this need to be equal to if $y_3(x)$ is a solution?

7. ## Re: Superposition principle-homogeneous equation

(c1+c2)y''+(c1+c2)y^2=0

8. ## Re: Superposition principle-homogeneous equation

If $y_3\equiv c_1y_1+c_2y_2$ is a solution, then we must have:

$y_3''+y_3^2=0$

Substitute for $y_3$ into the ODE using the above definition, and see if this is equal to your earlier result...

9. ## Re: Superposition principle-homogeneous equation

got it...thak you very much for guiding me...

10. ## Re: Superposition principle-homogeneous equation

The purpose of these problems is to show you that the linear superposition of solutions only forms a solution for linear equations.