Find out the answer to "Solving the ODEy′2−yy′+ex=0"

Deegan Chase

Answered question

2022-04-01

Solving the ODE

y^{' 2} - y y^{'} + e^{x} = 0

Answer & Explanation

Sawyer Anthony

Beginner2022-04-02Added 10 answers

Step 1
Use the substitution $y = 2 e^{\frac{x}{2}} v$
${\begin{cases} y = 2 e^{\frac{x}{2}} v \\ y^{'} = 2 e^{\frac{x}{2}} v^{'} + e^{\frac{x}{2}} v \end{cases} ⟹ e^{x} (2 v^{'} + v) (2 v^{'} - v) + e^{x} = 0$
or
$4 {v^{'}}^{2} - v^{2} + 1 = 0$
which can be solved by integrating directly
$\pm \frac{x}{2} + C = \int \frac{d v}{\sqrt{v^{2} - 1}} = {cosh}^{- 1} (v)$
which means
$v = \sqrt{1 + C^{2}} cosh (\frac{x}{2}) + C sinh (\frac{x}{2})$
and finally
$y = 2 \sqrt{1 + C^{2}} e^{\frac{x}{2}} cosh (\frac{x}{2}) + 2 C e^{\frac{x}{2}} sinh (\frac{x}{2})$
or
$y = (\sqrt{1 + C^{2}} + C) e^{x} + (\sqrt{1 + C^{2}} - C)$
Step 2
Even more simply, the original found answer can be simplified to
$y = C e^{x} + \frac{1}{C}$
he missing solutions can be found by going back to the equation
$4 {v^{'}}^{2} - v^{2} + 1 = 0$
The original solution assumed $v^{2} - 1 \neq 0$ so that we could divide by it later. In that case though, notice that $v = \pm 1$ is the constant solution
$4 {(0)}^{2} - {(\pm 1)}^{2} + 1 = 0$
so we also have
$y = \pm 2 e^{\frac{x}{2}}$
Moral of the story, always look for constant solutions in nonlinear DEs.

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