Solving a differential equation by using Laplace transform

Question

I need to solve this equations by using laplace-transform. I tried to solve it but when I reach to the point that it's needed to use partial fraction expansion in order to transform the laplace inverse, I get $\Delta<0$ and I don't know what to do next? because I always got $\Delta\ge0$, can you help me dear friends? $$y''+9y=\sin2t$$
$$y(0)=1$$ $$y'(0)=1$$ This is what I tried to do: $$L[y]=F(s);L[y']=S.F(s)-f(0);L[y'']=s^2.F(s) -s.f(0)-f'(0) \Rightarrow L[y'' + 9y] = L[\sin^2] \Rightarrow s^2.F(s)-s-1+9.F(s)={s^2\over s^2+4} \Rightarrow {2+(s^2+4)(s^2+1)\over (s^2+4)(s^2+9)} \Rightarrow $$ $${A\over s^2+4}+{B\over s^2+9}$$ that's it,here, I can't find $A,B$.

Answer 1

the charecteristic equation $$m^2+9=0$$ $$m_1,_2=\pm 3i$$ $$y_p=C_1\cos 3t+C_2\sin 3t$$ to find the particular soultion, we should change the $\sin^2t$ to $$\frac{1-\cos 2t}{2}$$ hence $$y''+9y=0.5-.5\cos 2t$$

we assume the particular solution as $$y_p=A+B\cos 2t+C\sin 2t$$

now you can complete the solution to find $A$,$B$, and $C$

Answer 2

The Laplace transform of $\sin{2t}$ is $2/(s^2 +4)$ instead of $s^2/(s^2 +4)$, so you are left with:

$$ F(s) = \frac{s^3+s^2+4 s+6}{\left(s^2+4\right) \left(s^2+9\right)} = \frac{B s + C}{s^2 + 4} + \frac{Ds + E}{s^2 + 9} = \frac{5 s+3}{5 \left(s^2+9\right)}+\frac{2}{5 \left(s^2+4\right)},$$ where I have used Mathematica to find out the corresponding coefficients. Inverse-Laplace transforming this should be easy. Can you take it from here? Do you need intermediate steps on finding the constants $B$, $C$,...?

Hope you find this helpful.

Cheers!

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Edit: snippet to compute the solution in Mathematica

In[1]:= ode = y''[x] + 9  y[x] == Sin[2 x];

In[2]:= odes = 
  LaplaceTransform[ode, x, s] /. {y[0] -> 1 , y'[0] -> 1, 
    LaplaceTransform[y[x], x, s] -> Fs};

In[3]:= F = Fs /. Solve[odes, Fs] [[1]] // Apart

Out[3]= 2/(5 (4 + s^2)) + (3 + 5 s)/(5 (9 + s^2))

In[4]:= InverseLaplaceTransform[F, s, x];

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Edit 2: as a remark, when you find a quadratic irreducible term in the denominator of the form $s^2 + k^2$ you should try a partial fraction decomposition of the form $(as + b)/(s^2 + k^2)$ instead of $a/(s^2 + k^2)$.

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Edit 3: how to obtain $B$, $C$, $D$ and $E$? We want:

$$ F(s) = \frac{\color{blue}{s^3+s^2+4 s+6}}{\left(s^2+4\right) \left(s^2+9\right)} = \frac{B s + C}{s^2 + 4} + \frac{Ds + E}{s^2 + 9} = \frac{\color{blue}{(Bs+C)(s^2 + 9) + (Ds + E)(s^2 + 4)}}{(s^2 + 4)(s^2+9)},$$ so the blue parts must coincide. The best way to do this is by matching up every power of $s$ in the numerator of LHS to the corresponding coefficient in the numerator of RHS, which, upon expanding, is given by:

$$ (B+D) s^3 + (C+E) s^2 + (9B + 4D) s + 9C+4E. $$ Therefore we require:

\begin{align} s^3: \quad & 1 = B+D, \\ s^2: \quad & 1 = C+E, \\ s^1: \quad & 1 = 9B + 4D, \\ s^0: \quad & 6 = 9C + 4E, \end{align} which hopefully yields the solution of your problem.