The below equation $(1)$ is second-ordinary differential equation, I have got the solution of it using standard method which is :$y(x)=e^{-3x}(-7+9e^x)$ , now my question is to solve equation $(1)$ using laplace transform :
$ y''+5y'+6y=0, y(0)=2, y'(0)=3\tag{1}$
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$$y''+5y'+6y=0, y(0)=2, y'(0)=3 \implies$$
$$[s^2F(s)-sy(0)-y'(0)]+5[sF(s)-y(0)]+6F(s)=0 \implies $$ $$(s^2+5s+6)F(s)=2s+13 \implies $$
$$F(s)=\frac{2s+13}{s^2+5s+6}=\frac{9}{s+2}-\frac {7}{s+3}\implies$$
$$ y(t) = 9e^{-2t}-7e^{-3t}$$
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this way dosent make things simple but it's still funny..
$$y''+5y'+6y=0, y(0)=2, y'(0)=3$$ $$y''+2y'+3y'+6y=0$$ Substitute $w=y'+2y$ $$w'+3w=0$$ $$\mathcal{ L}(w'+3w)=sF(s)-w(0)+3F(s)=0$$ Since $w(0)=y'(0)+2y(0)=7$ $$sF(s)-7+3F(s)=0$$ $$F(s)=\frac 7 {s+3}$$ $$w=7e^{-3x}$$ $$y'+2y=7e^{-3x}$$ $$\mathcal{L}(y'+2y)=7\mathcal{L}(e^{-3x})$$ $$sF(s)-y(0)+2F(s)=\frac 7 {s+3}$$ $$(s+2)F(s)=\frac 7 {s+3}+2$$ $$F(s)=\frac 7 {(s+2)(s+3)}+\frac 2 {s+2}=\frac 9 {s+2}-\frac 7 {s+3}$$ $$y(x)=9\mathcal{L^{-1}}(\frac 1 {s+2})-7\mathcal{L^{-1}}(\frac 1 {s+3})$$ $$\boxed{y(x)=9e^ {-2x}-7e^{-3x}}$$
Let $Y(s)$ be the laplace transform of $y(x)$. The first step is to take the Laplace transform of both sides. Recall that the Laplace transform is linear, and acts on derivatives in the following way: \begin{align*} \mathcal{L}(y')(s) &= sY(s) - y(0) \\ &= sY(s) - 2 \\ \mathcal{L}(y')(s) &= s^2Y(s) - sy(0) - y'(0) \\ &= s^2Y(s) - 2s - 3 \end{align*} Therefore, taking the Laplace transform of both sides yields, $$s^2Y(s) - 2s - 3 + 5(sY(s) - 2) + 6Y(s) = 0$$ Expanding and collecting terms of $Y(s)$ yields $$(s^2 + 5s + 6)Y(s) = 2s + 13 \implies Y(s) = \frac{2s + 13}{(s + 2)(s + 3)}.$$ Applying a partial fractions decomposition further gives us $$Y(s) = \frac{9}{s + 2} + \frac{-7}{s + 3}.$$ We have solved for the Laplace of $y(x)$, so we need to apply the inverse Laplace transform. Consulting a standard transform table, this gives us an answer of $$y(x) = 9e^{-2x} - 7e^{-3x}.$$