Find the general solution to: $$\frac{d^2y}{dx^2}+6\frac{dy}{dx}+13y=0$$ if $y(0)=3$ and $y'(0)=7$.
So I have tried solve it using Laplace transformation but cannot exceed from that....help me please
Here is what I have done so far:
$$s^2Y(s)-sy(0)+y'(0)+6sY(s)+6y(0)+13Y(s)=0$$ $$s^2Y(s)+7+6sY( s)+6(3)+13Y(s)=0$$ $$(s^2 +6s+13)Y(s)=25$$ $$Y(s)=\frac{25}{s^2 +6s+13}$$
You seem to have made some sign mistakes: $$s^2Y(s)-sy(0)\color{red}{-}y'(0)+6sY(s)\color{red}{-}6y(0)+13Y(s)=0$$ Hence, given your initial conditions, you should obtain: $$s^2 Y(s)-3s-7+6sY(s)-18+13Y(s)=0$$ $$Y(s)=\frac{25+3s}{s^2+6s+13}$$ Hence, the problem boils down to evaluating the RHS: $$y(x)=\mathcal{L}^{-1}\left\{\frac{25+3s}{s^2+6s+13}\right\}$$
To evaluate the Inverse Laplace Transform of $Y(s)$ in a convenient manner, we complete the square on the denominator and then write $s=(s+3)-3$ for the numerator. We then split the fraction in two, giving: $$Y(s)=\frac{25+3s}{(s+3)^2+4}=\color{red}{\frac{16}{(s+3)^2+4}}+\color{green}{\frac{3(s+3)}{(s+3)^2+4}} \tag{1}$$
You can now evaluate the Inverse Laplace Transform of $Y(s)$ by applying the following well-known results: (Results 19 and 20 from the Table of Laplace Transforms) $$\mathcal{L}\left\{e^{ax}\sin(bx)\right\}=\frac{b}{(s-a)^2+b^2} \tag{2.1}$$ $$\mathcal{L}\left\{e^{ax}\cos(bx)\right\}=\frac{s-a}{(s-a)^2+b^2} \tag{2.2}$$ If you want a derivation of $(2.1)$, I derived it on this answer. The derivation of $(2.2)$ is very similar.