# Thread: 10.09.14t Solve the differential equation

1. ## 10.09.14t Solve the differential equation

$\textrm{Solve the differential equation}$
\begin{align*}\displaystyle
e^x\frac{dy}{dx}+5e^x y&=4, x>0 \\
&=
\end{align*}

confused about the $\frac{dy}{dx}$

2. ## Re: 10.09.14t Solve the differential equation

Originally Posted by bigwave
$\textrm{Solve the differential equation}$
\begin{align*}\displaystyle
e^x\frac{dy}{dx}+5e^x y&=4, x>0 \\
&=
\end{align*}

confused about the $\frac{dy}{dx}$
one way to go about it is

$e^x \dfrac{dy}{dx} + 5 e^x y = 4$

$\dfrac {dy}{dx} + 5 y = 4e^{-x}$

solving the homogeneous equation

$\dfrac{dy}{dx} = -5y$

$y = c_1 e^{-5x}$

let $y_p = c_2 e^{-x}$

$-e^{-x}+5 c_2 e^{-x} = 4 e^{-x}$

$5c_2 - 1 = 4$

$c_2 = 1$

$y(x) = c_1 e^{-5x} + e^{-x}$

3. ## Re: 10.09.14t Solve the differential equation

When the equation is first order linear, it's usually expected that the student should use an integrating factor.

\displaystyle \begin{align*} \frac{\mathrm{d}y}{\mathrm{d}x} + 5\,y &= 4\,\mathrm{e}^{-x} \end{align*}

The integrating factor is \displaystyle \begin{align*} \mathrm{e}^{\int{5\,\mathrm{d}x}} = \mathrm{e}^{5\,x} \end{align*}, so multiplying both sides by the integrating factor gives

\displaystyle \begin{align*} \mathrm{e}^{5\,x}\,\frac{\mathrm{d}y}{\mathrm{d}x} + 5\,\mathrm{e}^{5\,x}\,y &= \mathrm{e}^{4\,x} \\ \frac{\mathrm{d}}{\mathrm{d}x}\,\left( \mathrm{e}^{5\,x}\,y \right) &= \mathrm{e}^{4\,x} \\ \mathrm{e}^{5\,x}\,y &= \int{ \mathrm{e}^{4\,x} \,\mathrm{d}x} \\ \mathrm{e}^{5\,x}\,y &= \frac{1}{4}\,\mathrm{e}^{4\,x} + C \\ y &= \frac{1}{4}\,\mathrm{e}^{-x} + C\,\mathrm{e}^{-5\,x} \end{align*}

4. ## Re: 10.09.14t Solve the differential equation

The first answer is correct, the second is not. Prove It is correct that the "integrating factor" is $e^{5x}$ but he lost a "4" when he multiplied both sides of the equation by that.