zeotropojd

2021-12-26

Solve the following differential equations for a general solution: $y3{y}^{\prime }+2y={e}^{x}$

Kayla Kline

Given differential equation:
$y3{y}^{\prime }+2y={e}^{x}$ (1)
$y3{y}^{\prime }+2y=0$ (2)
The characteristic equation of (2) is
${m}^{2}+3m+2=0$
$m=\frac{-3±\sqrt{9-8}}{2}$
$m=\frac{-3±1}{2}$

therefore general solution of (2)
$y\left(x\right)={c}_{1}{e}^{-x}+{c}_{2}{e}^{-2x}$
If we assume a particular solution of non homogeneous equation is of the form.
${y}_{p}\left(x\right)=k{e}^{x}$
then ${y}_{p}=k{e}^{x},{y}_{p}k{e}^{x}$
Putting these values in (1), we have
$k{e}^{x}+3\left(k{e}^{x}\right)+2\left(k{e}^{x}\right)={e}^{x}$
$⇒\left(k+2k+3k\right){e}^{x}={e}^{x}$
$⇒\left(k+2k+3k\right)=1$
$⇒6k=1$
$⇒k=\frac{1}{6}$
therefore ${y}_{p}=\frac{1}{6}{e}^{x}$.
Hence: General solution of equation (1) is
$y\left(x\right)={c}_{1}{e}^{-x}+{c}_{2}{e}^{-2x}+\frac{1}{6}{e}^{x}$ where are constant.

David Clayton

$y=\mathrm{exp}\left(k\cdot x\right)$
${k}^{2}-3\cdot k+2=0$
${k}_{1}=2$
${k}_{2}=1$
$y=C1\cdot \mathrm{exp}\left(x\right)+C2\cdot \mathrm{exp}\left(2\cdot x\right)$
$y=C1\cdot \mathrm{exp}\left(x\right)+C2\cdot \mathrm{exp}\left(2\cdot x\right)-x\cdot \mathrm{exp}\left(x\right)$

karton

Complimentary function
Auxiliary equation ${r}^{2}+3r+2=0⇒\left(r+2\right)\left(r+1\right)=0⇒r=-2,-1$
${y}_{c}=A{e}^{-x}+B{e}^{-2x}$
Particular Integral
$\left(D+1\right)\left(D+2\right)y={e}^{ex};\text{Let}\left(D+2\right)y=z$
$\left(D+1\right)={e}^{ex}$
$\frac{dz}{dx}+z={e}^{ex}$
$I.F.={e}^{\int dx}={e}^{x}$
Multiplying by I.F. ${e}^{x}z=\int {e}^{x}{e}^{ex}dx$
Put ${e}^{x}=t;{e}^{x}dx=dt$
${e}^{x}z=\int {e}^{t}dt={e}^{t}={e}^{ex}$
$z={e}^{-x}{e}^{ex}$
$\left(D+2\right)y={e}^{-x}{e}^{ex}$
$\frac{dy}{dx}+2y={e}^{-x}{e}^{ex}$
$I.F={e}^{\int 2dx}={e}^{2x}$
Multiplying by I.F. and integrating
${e}^{2x}y=\int {e}^{2x}{e}^{-x}{e}^{ex}dx=\int {e}^{x}{e}^{ex}dx={e}^{ex}$ as above.
So ${y}_{p}={e}^{-2x}{e}^{ex}$
The total solution is $y=A{e}^{-x}+B{e}^{-2x}+{e}^{-2x}{e}^{ex}$

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