Find the general solution of the differential equation
yy''+2\left(y\right)^2=0
Solution:
Based on Special Second-Ordered Differential Equation: Special case 3
F\left(\frac{d^2y}{dx^2},\:\frac{dy}{dx},\:y\right)=0
Denote and substitute to the given equation.
P= y' =\frac{dy}{dx} \\ P\frac{dp}{dy}= y'' =\frac{d^2y}{dx^2}
We will have,
y(P\frac{dp}{dx})+2(P)^2=0
Divide both sides with
\:\frac{1}{yP}
We will come to,
\frac{dp}{dy}+\frac{2P}{y}=0
Tranpose,
\frac{2P}{y}
We will have
\frac{dp}{dy}=-\frac{2P}{y}
Integrate both sides,
\int \frac{dp}{dy}=-\int\frac{2P}{y}
The equation will become a SEPARABLE DIFFERENTIAL EQUATION, multiply both sides with
\frac{dy}{P}\:
We will come to the equation:
\frac{dp}{P}=-\frac{2}{y}dy
Integrate both sides,
\int \frac{dp}{P}=-\int\frac{2}{y}dy
The answer will be:
\ln \left(P\right)=\ln \left(y^{-2}\right)+lnC
Apply logarithmic definition and exponent rule
loga^b=c\:then,\:b=a^c\\a^{b+c}=a^ba^c
The answer will be:
P=\frac{C}{y^2}
Recall that
P=\frac{dy}{dx}
Substitute the original value of P,
\frac{dy}{dx}=\frac{C}{y^2}
Again, this is a Separable Differential Equation, multiply both sides with:
y^{2}dx
It will become
y^{2}dy=Cdx
Integrate both sides,
\int y^{2}dy=\int Cdx
The answer will be
\frac{y^3}{3}=C1x+C2
Multiply both sides with 3 and the final answer will be
y^3=C_1x+C_2
You can still solve it explicitly,
y=\sqrt[3]{C_1x+C_2}
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