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Please help me figure out the right side of the equation. Find the general solution of the following equations.
$$y''-2y'+2y=e^x\sin x$$ .
I know that the general solution is
$$y=e^x(c_1\cos x+c_2\sin x)$$
But I am confused about the particular solution due to the combination of the exponential and sine terms.
I know that individually, $e^x$ will yield $y_p=Ae^{ax}$ and $\sin x$ will yield $y_p=A\sin bx + B \cos bx$ but I don't know how to combine these two. Also, if the combination of these two is the product of their particular solutions, then we fact another problem. Since we can't simply do $$y_p=Ae^{ax}(A\sin bx+B \cos bx)$$ since this looks like our general solution, perhaps we need to multiply this expression by an x, which will yield $$y_p=(x)Ae^{ax}(A\sin bx+B \cos bx)$$. Please let me know if I am on the correct track and help me finish the problem.

I know that the final answer is $$y_p=-\frac{1}{2}xe^x\cos x$$ but I am not able to obtain it.

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  • $\begingroup$ Thank you. Also, I made a mistake, the answer is negative. $\endgroup$ – thisisme Jul 4 '17 at 3:04
  • $\begingroup$ When I solved the problem given your equation I obtained $y_p=e^x-2/3\cos x+-1/3\sin x$ which is not the same as the back of the book. Could you please help me find my mistake? $\endgroup$ – thisisme Jul 4 '17 at 3:27
  • $\begingroup$ So sorry for the confusion, I initially had typos in my problem. The original question is the product as I have above. Using your logic, would the particular solution be, $y_p=ae^x(b\cos x+c\sin x)$? $\endgroup$ – thisisme Jul 4 '17 at 3:30
  • $\begingroup$ I see, thank you and sorry for the confusion. Is the reason you have the x so that it does not look like the general solution? $\endgroup$ – thisisme Jul 4 '17 at 3:32
  • $\begingroup$ I now understand, thanks a lot for your help! $\endgroup$ – thisisme Jul 4 '17 at 3:34
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Consider the following second order linear homogeneous differential equation $$\frac{d^2y}{dx^2}-2\frac{dy}{dx}+2y=0$$ Note that the coefficients are constant! Finding the complementary function is then $$m^2-2m+2=0$$

This would be a complex roots since its discriminant is $<0$! By method of quadratic formula we have $$m=\frac{-(-2) \pm \sqrt{4-4(1)(2)}}{2(1)}$$

$$m=1\pm i$$ The complementary function is then $$y_c=e^x(c_1cosx+c_2sinx)$$

Now method of undetermined coefficients

$$S_1={e^x}$$ Did it repeat inside the complementary function the answer is no!!!

$$S_2={sinx,cosx}$$

Did it repeat inside the complementary function or other set.

SO we have the following particular solution

$$y_p=Ae^x+Bcosx+Csinx$$

I hope that you can continue to seek out the undetermined coefficients from this step on !

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