Oliver Tyler

2022-03-28

Determine the general solution of the given differential equation:
$y{}^{″}-4{y}^{\prime }+13y={e}^{2x}$

allemelrypi0e

Step 1
$y\left(x\right)={y}_{h}\left(x\right)+{y}_{p}\left(x\right),$
where ${y}_{h}\left(x\right)$ is a solution of a homogeneous equation $y{}^{″}-4{y}^{\prime }+13y=0$, and ${y}_{p}\left(x\right)$ is a particular solution.
To solve the equation $y{}^{″}-4{y}^{\prime }+13y=0$, one have to solve a characteristic equation
${\lambda }^{2}-4\lambda +13=0,$
which is a quadratic algebraic equation. The solution is
${\lambda }_{1,2}=\frac{-\left(-4\right)±\sqrt{{\left(-4\right)}^{2}-4\cdot 1\cdot 13}}{2\cdot 1}=2±3i⇒$
${y}_{h}\left(x\right)={C}_{1}{e}^{2x}\mathrm{cos}\left(3x\right)+{C}_{2}{e}^{2x}\mathrm{sin}\left(3x\right)$.
Step 2
The solution ${y}_{p}\left(x\right)$ can be found in a form ${y}_{p}\left(x\right)=A{e}^{2x}$. Then,

Substituting into the equation, one has
$4A{e}^{2x}-4\cdot 2A{e}^{2x}+13A{e}^{2x}={e}^{2x}⇔9A=1⇔A=\frac{1}{9},$
and ${y}_{p}\left(x\right)=\frac{1}{9}{e}^{2x}$
Finally, the general solution is
$y\left(x\right)={y}_{h}\left(x\right)+{y}_{p}\left(x\right)={C}_{1}{e}^{2x}\mathrm{cos}\left(3x\right)+{C}_{2}{e}^{2x}\mathrm{sin}\left(3x\right)+\frac{1}{9}{e}^{2x}$.

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