How find this equation $y''+(y')^2\cdot e^x=0$

Question

if the ODE $$y''+(y')^2\cdot e^x=0$$ such $$y(0)=1,y'(0)=1$$

Find the $y(x)=?$

my ugly methods: let $$y'=p,y''(x)=p'(x)$$ so $$p'(x)+p^2\cdot e^x=0$$ $$\dfrac{dp}{dx}=-p^2e^x$$ $$\dfrac{dp}{-p^2}=e^xdx$$ $$\Longrightarrow \dfrac{1}{p}=e^x+c\Longrightarrow p=e^{-x}(p(0)=1)$$ so $$y(x)=2-e^{-x}$$

My question: This ODE have other methods,? such I want use $$(y'\cdot e^x)'=y''e^x+e^xy'$$ Thank you

Answer 1

My first thought when I saw your questions was that it is non-linear differentiation and there's an $e^x$ term; as such, I would try $$y(x) = e^{k x}$$ where $k$ is such that (by considering the powers/primes in the ODE) $$k + 2k + 1 = 0 \Rightarrow k = -1.$$ This then gives the same result as yours (after using the initial condition).

Also a good idea would be to let, for example, $Y = y'$, and solve the 1st order ODE for $Y$, then integrate for $y'$.

Hope this helps! If so, please remember to upvote and/or accept this answer! =D

Answer 2

Rewrite: $$ \begin{align} y''+(y')^2e^x&=0\\ y''&=-(y')^2e^x\\ \frac{d(y')}{dx}&=-(y')^2e^x\\ -\frac{d(y')}{(y')^2}&=e^x\ dx\\ -\int\frac{d(y')}{(y')^2}&=\int e^x\ dx\\ \frac{1}{y'}&=e^x+C_1\\ y'&=\frac{1}{e^x+C_1}\\ \frac{dy}{dx}&=\frac{1}{e^x+C_1}\\ \int\ dy&=\int \frac{1}{e^x+C_1}\ dx\\ &=\int \frac{e^{-x}}{1+C_1e^{-x}}\ dx. \end{align} $$ Now, let $u=1+C_1e^{-x}$ and $du=-C_1e^{-x}\ dx$, then \begin{align} \int\ dy&=\int \frac{e^{-x}}{1+C_1e^{-x}}\ dx\\ \int\ dy&=\int \frac{e^{-x}}{u}\cdot \frac{du}{-C_1e^{-x}}\\ \int\ dy&=-\frac{1}{C_1}\int \frac{du}{u}\\ y&=-\frac{1}{C_1}\ln (1+C_1e^{-x})+C_2\\ y&=\frac{x-\ln(e^x+C_1)}{C_1}+C_2. \end{align}

Answer 3

Rewrite: $$ \begin{align} y''+(y')^2e^x&=0\\ y''&=-(y')^2e^x\\ \frac{d(y')}{dx}&=-(y')^2e^x\\ -\frac{d(y')}{(y')^2}&=e^x\ dx\\ -\int\frac{d(y')}{(y')^2}&=\int e^x\ dx\\ \frac{1}{y'}&=e^x+C_1\\ y'(x)&=\frac{1}{e^x+C_1}\\ \end{align} $$ For $y'(0)=1$, we get $C_1=0$ and $y'(x)=e^{-x}$. Hence \begin{align} \frac{dy}{dx}&=e^{-x}\\ \int\ dy&=\int e^{-x}\ dx\\ y(x)&=-e^{-x}+C_2 \end{align} For $y(0)=1$, we get $C_2=2$ and $y(x)=2-e^{-x}$.

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Credit answer : Mr. Tunk-Fey (>‿◠)✌