I want to solve $$y''+y=x+1,y(0)=1$$ My solution:

I solve he differential equation and I get $$y=c_1\cos x+c_2\sin x+x+1$$ Then I know that $y(0)=1\implies c_1=0$

Finally I get $$\boxed{y=c\sin x+x+1}$$

Wolfram solution:

if I type in wolfram solver $y''+y=x+1,y(0)=1$ and the result I get is $$\boxed{y=c\sin x}$$

Which one is the correct solution finally?

  • $\begingroup$ $y(0) = c \sin(0) = 0 \neq 1$ $\endgroup$ – RayX Sep 26 '15 at 16:37
  • $\begingroup$ At least we can tell that the last solution doesn't satisfy the initial condiition. $\endgroup$ – pointer Sep 26 '15 at 16:37
  • $\begingroup$ you need a second boundary condition... $\endgroup$ – tired Sep 26 '15 at 16:38
  • 2
    $\begingroup$ Moral of the story: trust your own mathematics and logic. Your solution is correct. $\endgroup$ – Simon S Sep 26 '15 at 16:46
  • 1
    $\begingroup$ Completing the query to "y''+y=x+1,y(0)=1, y'(0)=a" gives again the correct solution. $\endgroup$ – LutzL Sep 26 '15 at 16:52

Your Answer:

$y''+y=-c\sin x+c\sin x+x+1=x+1$

Also $y(0)=1$. The same cannot be said about WA's answer.


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