Elementary Differential Equations by Dela Fuente, Feliciano and Uy Chapter 9 Problem 5 — Special Second-Ordered Differential Equations

Find the general solution of the differential equation

yy′′+2(y)2=0yy''+2\left(y\right)^2=0

Solution:

Based on Special Second-Ordered Differential Equation: Special case 3

F(d2ydx2, dydx, y)=0F\left(\frac{d^2y}{dx^2},\:\frac{dy}{dx},\:y\right)=0

Denote and substitute to the given equation.

P=y′=dydxPdpdy=y′′=d2ydx2P= y' =\frac{dy}{dx} \\ P\frac{dp}{dy}= y'' =\frac{d^2y}{dx^2}

We will have,

y(Pdpdx)+2(P)2=0y(P\frac{dp}{dx})+2(P)^2=0

Divide both sides with

 1yP \:\frac{1}{yP}

We will come to,

dpdy+2Py=0\frac{dp}{dy}+\frac{2P}{y}=0

Tranpose,

2Py\frac{2P}{y}

We will have

dpdy=−2Py\frac{dp}{dy}=-\frac{2P}{y}

Integrate both sides,

∫dpdy=−∫2Py\int \frac{dp}{dy}=-\int\frac{2P}{y}

The equation will become a SEPARABLE DIFFERENTIAL EQUATION, multiply both sides with

dyP \frac{dy}{P}\:

We will come to the equation:

dpP=−2ydy \frac{dp}{P}=-\frac{2}{y}dy

Integrate both sides,

∫dpP=−∫2ydy\int \frac{dp}{P}=-\int\frac{2}{y}dy

The answer will be:

ln⁡(P)=ln⁡(y−2)+lnC\ln \left(P\right)=\ln \left(y^{-2}\right)+lnC

Apply logarithmic definition and exponent rule

logab=c then, b=acab+c=abacloga^b=c\:then,\:b=a^c\\a^{b+c}=a^ba^c

The answer will be:

P=Cy2P=\frac{C}{y^2}

Recall that

P=dydxP=\frac{dy}{dx}

Substitute the original value of P,

dydx=Cy2\frac{dy}{dx}=\frac{C}{y^2}

Again, this is a Separable Differential Equation, multiply both sides with:

y2dxy^{2}dx

It will become

y2dy=Cdxy^{2}dy=Cdx

Integrate both sides,

∫y2dy=∫Cdx\int y^{2}dy=\int Cdx

The answer will be

y33=C1x+C2\frac{y^3}{3}=C1x+C2

Multiply both sides with 3 and the final answer will be

y3=C1x+C2y^3=C_1x+C_2

You can still solve it explicitly,

y=C1x+C23y=\sqrt[3]{C_1x+C_2}
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