Solve the differential equation (1 + y2e2x) y′ + y = 0

Consider the differential equation

$(1+y^2 e^{2x})y' + y = 0.$

Make a change of variable $y = ue^{mx}$ where $u$ is a function of $x$ and $m$ is a constant and solve the differential equation.

Incomplete.

April 20, 2019 at 1:12 am

Evangelos says:

We have the equation

$\begin{align*} (1 + y^{2}e^{2x})y' + y &= 0 \end{align*}$

Which can be re-written as

$\begin{align*} y' &= \frac{-y}{1 + y^{2}e^{2x}} \end{align*}$

Now, we use our substitution of variable

$\begin{align*} y &= ue^{mx} \\ y' &= u'e^{mx} + mue^{mx} \end{align*}$

Giving us

$\begin{align*} u'e^{mx} + mue^{mx} &= \frac{-ue^{mx}}{1 + u^{2}e^{2(m+1)x}} \end{align*}$

Cancelling out the exponential on each side gives us

$\begin{align*} u' + mu &= \frac{-u}{1 + u^{2}e^{2(m+1)x}} \end{align*}$

Now, a small change in notation to make the rest more intuitive. Using Leibniz notation instead of prime notation, as u and u’ are functions of x.

$\begin{align*} \frac{du}{dx} + mu &= \frac{-u}{1 + u^{2}e^{2(m+1)x}} \end{align*}$

If we set m = -1, the equation becomes separable

$\begin{align*} \frac{du}{dx} + -u &= \frac{-u}{1 + u^{2}} \end{align*}$

$\begin{align*} \frac{du}{dx} &= \frac{-u}{1 + u^{2}} + u \\ &= \frac{u^{3}}{1 + u^{2}} \end{align*}$

Separating variables and integrating both sides gives us

$\begin{align*} \int \frac{1 + u^{2}}{u^3} du &= \int dx \\ -\frac{1}{2}u^{-2} + \log|u| &= x + C \end{align*}$

Now, we can substitute u as follows

$\begin{align*} y &= ue^{mx} \\ u &= ye^{-mx} \\ &= ye^x \end{align*}$

$\begin{align*} -\frac{1}{2}y^{-2}e^{-2x} + \log|ye^x| &= x + C \end{align*}$

Simplifying terms gives us

$\begin{align*} y^{2}\log|y| &= \frac{1}{2}e^{-2x} + Cy^2 \end{align*}$

Which is our back-of-book answer.

Stumbling Robot