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I want to guess for the particular solution to the given differential equation without finding the coefficients.

$4y"+y=e^{-2t}\sin{(\frac{t}{2}})$+ 6t $\cos(\frac{t}{2}).$

The guess for the form of the particular solution is

$Y_p(t)=e^{-2t}\big(A\cos{(\frac{t}{2})}+B\sin{(\frac{t}{2})}\big)+t(Ct+D)\cos{(\frac{t}{2})}+t(Et+f)\sin{(\frac{t}{2})}$

But the complementary solution to this differential equation is

$y_c(t)=c_1\cos{(\frac{t}{2})}+c_2\sin{(\frac{t}{2})}$.

I want to know how the complementary solution is computed? If any one knows the answer, post the reply.

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$$4y"+y=e^{-2t}\sin(t/2)+6t\cos(t/2)$$ The characteristic equation is $$4r^2+1=0 \implies r^2-\frac {i^2}4=0 $$$$\implies (r-i/2)(r+i/2)=0 \implies r=\pm\frac i2$$ Therefore $$y(t)=c_1\cos(t/2)+c_2\sin(t/2)$$

Your guess is correct for the particular solution $$4y"+y=e^{-2t}\sin(t/2)$$ $$ \implies y_p=e^{-2t}(A \cos(t/2)+B\sin(t/2))$$ $$4y"+y=6t\cos(t/2)$$ $$\implies y_p=t(Ct+D)\cos(t/2)+t(Et+F)\sin(t/2)$$

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  • $\begingroup$ Complex number definition is $\sqrt{-a}=\pm i\sqrt{a}$ $\endgroup$ – Dhamnekar Winod Oct 18 '18 at 17:09
  • $\begingroup$ @DhamnekarWinod here you have that $$r^2-\frac {i^2}4= 0 \implies (r-\frac i2)(r+\frac i2)=0 \implies r=\pm i/2$$ $\endgroup$ – Isham Oct 18 '18 at 17:18
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    $\begingroup$ The characteristic equation is $4r^2+1=0$ divided by the whole equation by 4, we get $r^2+ \frac14=0,\rightarrow r^2= -\frac14, \rightarrow r=\sqrt{-\frac14}$ and $\sqrt{-\frac14}=\pm \frac12i$ $\endgroup$ – Dhamnekar Winod Oct 18 '18 at 17:24
  • $\begingroup$ @DhamnekarWinod you can do the calculation that way too...$r=\pm \sqrt {-1/4}$ but remember that we have also $a^2-b^2=(a-b)(a+b)$ $\endgroup$ – Isham Oct 18 '18 at 17:24
  • $\begingroup$ Your method of calculation is also correct. But i understood it lately. $\endgroup$ – Dhamnekar Winod Oct 18 '18 at 17:27

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