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

Question
Equations
asked 2021-01-02
Solve. \(\displaystyle{y}\text{}{\left({y}'\right)}^{{2}}={2}{e}^{{-{y}}}\)

Answers (1)

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:\(\displaystyle\frac{{1}}{{2}}\sqrt{{{C}+{4}{e}{e}^{{y}}}}=\pm{x}+{D}\)
0

Relevant Questions

asked 2021-01-16
Solve the following systems of equations.
\(\displaystyle{x}^{{2}}+{y}^{{2}}−{19}{x}+{y}+{66}={0}\)
x−y−3=0
asked 2020-12-09
Solve the system of equations.
3(x-y)=y-3
\(\displaystyle-{2}{x}+\frac{{y}}{{2}}=-{11}\)
asked 2021-02-18
Consider the following system of llinear equations.
\(\displaystyle\frac{{1}}{{3}}{x}+{y}=\frac{{5}}{{4}}\)
\(\displaystyle\frac{{2}}{{3}}{x}-\frac{{4}}{{3}}{y}=\frac{{5}}{{3}}\)
Part A: \(\displaystyle\frac{{{W}\hat{\propto}{e}{r}{t}{y}}}{{\propto{e}{r}{t}{i}{e}{s}}}\) can be used to write an equivalent system?
Part B: Write an equivalent system and use elimination method to solve for x and y.
asked 2020-11-30
Solve the following system of Equations.
\(\displaystyle-{2}{x}^{{2}}-{y}^{{2}}=-{200}\)
\(\displaystyle-{3}{x}^{{2}}-{y}^{{2}}=-{300}\)
asked 2020-12-22
Solve the system of equations: \(\displaystyle{3}{x}-{y}=-{2},{2}{x}^{{2}}-{y}={0}\)
asked 2020-10-27
Solve the following system of equations.
\(\displaystyle{2}{x}-\frac{{1}}{{5}}{y}=\frac{{4}}{{5}}\)
\(\displaystyle\frac{{1}}{{3}}{x}-\frac{{1}}{{2}}{y}=-{12}\)
x-?
y-?
asked 2020-10-18
Use the method of substitution to solve the following system of equations. If the system is dependent, express the solution set in terms of one of the variables. Leave all fractional answers in fraction form.
y=-11
x+2y=7
asked 2020-10-20
System of equations. Use matrices to solve
2x+y=-10
6x-3y=6
asked 2021-02-04
Solve the system of equations.
4x+y=16
2x+3y=18
asked 2021-02-05
Use Cramer’s Rule to solve (if possible) the system of linear equations.
4x-y-z=1
2x+2y+3z=10
5x-2y-2z=-1
...