# Solve. y"+(y')^2=2e^(-y)

Question
Equations
Solve. $$\displaystyle{y}\text{}{\left({y}'\right)}^{{2}}={2}{e}^{{-{y}}}$$

2021-01-03
Step 1
We do not have x in this equation, so the substitution is
p(y)=y',
where $$\displaystyle{y}'=\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}$$.So,
$$\displaystyle{y}\text{}\frac{{{\left.{d}{y}\right.}'}}{{{\left.{d}{x}\right.}}}=\frac{{{d}{p}}}{{{\left.{d}{x}\right.}}}=^{{\cdot}}\frac{{{d}{p}}}{{{\left.{d}{y}\right.}}}\cdot\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}={p}'{y}'={p}'{p}$$,
where in (*) we used the Chain Rule, and p' is a derivative of p with respect to y.
So, the initial equation becomes
$$\displaystyle{p}'{p}+{p}^{{2}}={2}{e}^{{-{y}}}$$...(1)
Notice that this is a Bernoulli equation, with substitution
$$\displaystyle{u}={p}^{{2}}\Rightarrow{u}'={2}{p}{p}'\Rightarrow{p}{p}'=\frac{{{u}'}}{{2}}$$
Therefore, (1)becomes
$$\displaystyle\frac{{{u}'}}{{2}}+{u}={2}{e}^{{-{y}}}\Rightarrow{u}'+{2}{u}={4}{e}^{{-{y}}}$$...(2)
We first solve the homogeneous equation
u'+2u=0
The characteristic equation is
$$\displaystyle{r}+{2}={0}\Rightarrow{r}=-{2}$$
Therefore, the solution to homogeneous equation is
$$\displaystyle{u}={C}{e}^{{-{2}{y}}}$$
Now we need to find the particular solution of (2).We will try with $$\displaystyle{u}={D}{e}^{{-{y}}}$$:
$$\displaystyle{u}'+{2}{u}={4}{e}^{{-{y}}}\Rightarrow-{D}{e}^{{-{y}}}+{2}{D}{e}^{{-{y}}}={4}{e}^{{-{y}}}\Rightarrow{D}={4}$$
Step 2
Therefore, the solution of(2) is
$$\displaystyle{u}={C}{e}^{{-{2}{y}}}+{4}{e}^{{-{y}}}$$
Therefore, since $$\displaystyle{u}={p}^{{2}}$$, so $$\displaystyle{p}=\pm\sqrt{{{u}}}$$ we get
$$\displaystyle{p}=\pm\sqrt{{{C}{e}^{{-{2}{y}}}+{4}{e}^{{-{y}}}}}$$
Now recall that p = y', so
$$\displaystyle{y}'=\pm\sqrt{{{C}{e}^{{-{2}{y}}}+{4}{e}^{{-{y}}}}}\Rightarrow\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}=\pm\sqrt{{{C}{e}^{{-{2}{y}}}+{4}{e}^{{-{y}}}}}$$
This is a separable equation, so
$$\displaystyle\frac{{{\left.{d}{y}\right.}}}{{\sqrt{{{C}{e}^{{-{2}{y}}}+{4}{e}^{{-{y}}}}}}}=\pm{\left.{d}{x}\right.}\Rightarrow\int\frac{{{\left.{d}{y}\right.}}}{{\sqrt{{{C}{e}^{{-{2}{y}}}+{4}{e}^{{-{y}}}}}}}=\pm\int{\left.{d}{x}\right.}$$...(3)
The RHS integral is easy:
$$\displaystyle\pm\int{\left.{d}{x}\right.}=\pm{x}+{D}_{{2}}$$,
where $$\displaystyle{D}_{{2}}$$ is a constant.
Step 3
On the other hand,
$$\displaystyle\int\frac{{{\left.{d}{y}\right.}}}{{\sqrt{{{C}{e}^{{-{2}{y}}}+{4}{e}^{{-{y}}}}}}}=\int\frac{{{\left.{d}{y}\right.}}}{{{e}^{{-{y}}}\sqrt{{{C}+{4}{e}^{{y}}}}}}$$
$$\displaystyle=\int\frac{{{e}^{{y}}{\left.{d}{y}\right.}}}{{\sqrt{{{C}+{4}{e}^{{y}}}}}}$$
$$\displaystyle={\left\lbrace{\left({u}={e}^{{y}}\right)},{\left({d}{u}={e}^{{y}}{\left.{d}{y}\right.}\right)}\right\rbrace}$$
$$\displaystyle=\int\frac{{{d}{u}}}{{\sqrt{{{C}+{4}{u}}}}}$$
$$\displaystyle=\int{\left({C}+{4}{u}\right)}^{{-\frac{{1}}{{2}}}}{d}{u}$$
$$\displaystyle=\frac{{1}}{{4}}\frac{{{\left({C}+{4}{u}\right)}^{{\frac{{1}}{{2}}}}}}{{\frac{{1}}{{2}}}}+{D}_{{1}}$$
$$\displaystyle\frac{{1}}{{2}}\sqrt{{{C}+{4}{u}}}+{D}_{{1}}$$
$$\displaystyle=\frac{{1}}{{2}}\sqrt{{{C}+{4}{e}^{{y}}}}+{D}_{{1}}$$,
where $$\displaystyle{D}_{{1}}$$ is some constant. So, after defining $$\displaystyle{D}={D}_{{2}}-{D}_{{1}}$$, (3) becomes
$$\displaystyle\frac{{1}}{{2}}\sqrt{{{C}+{4}{e}{e}^{{y}}}}=\pm{x}+{D}$$
Result:$$\displaystyle\frac{{1}}{{2}}\sqrt{{{C}+{4}{e}{e}^{{y}}}}=\pm{x}+{D}$$

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