Second-Order Reducible DEs

Second-Order Reducible DE
These have the form F(x, y, y′, y′′)=0F\left(x,\ y,\ y',\ y''\right)=0F(x, y, y′, y′′)=0
Case 1: Dependent Varible y is Missing
If our DE is of the form F(x, y′, y′′)=0F\left(x,\ y',\ y''\right)=0F(x, y′, y′′)=0, then 
we let v=y′v=y'v=y′ and v′=y′′v'=y''v′=y′′
solve the DE for the solution v=...v=...v=...
solve for the solution y=...y=...y=... using v=dydxv=\frac{dy}{dx}v=dxdy​

Example:
Solve the DE xy′′+y′=2xxy''+y'=2xxy′′+y′=2x

1. Let v=y′v=y'v=y′ and v′=y′′v'=y''v′=y′′ to get xv′+v=2xxv'+v=2xxv′+v=2x

2. Now we can solve this like a DE in terms of vvv abd xxx
x dvdx+v=2xx\ \frac{dv}{dx}+v=2xx dxdv​+v=2x
dvdx+v(1x)=2\frac{dv}{dx}+v\left(\frac{1}{x}\right)=2dxdv​+v(x1​)=2

This is a linear DE:
P=1xP=\frac{1}{x}P=x1​
Q=2Q=2Q=2
Integrating factor R=e∫1xdx=eln⁡x=x\displaystyle R=e^{\int\frac{1}{x} dx}=e^{\ln x}=xR=e∫x1​dx=elnx=x
v=1x∫2xdxv=\frac{1}{x}\int_{ }^{ }2xdxv=x1​∫​2xdx
v=1x[x2+C]v=\frac{1}{x}\left[x^2+C\right]v=x1​[x2+C]
v=x+Cxv=x+\frac{C}{x}v=x+xC​

3. Finally since dydx=v\frac{dy}{dx}=vdxdy​=v:
dydx=x+Cx\frac{dy}{dx}=x+\frac{C}{x}dxdy​=x+xC​
y=∫x+Cxdxy=\int_{ }^{ }x+\frac{C}{x}dxy=∫​x+xC​dx
y=x22+Cln⁡∣x∣+D\boxed{y=\frac{x^2}{2}+C\ln |x|+D}y=2x2​+Cln∣x∣+D​

*You can always do a check by substituting y′y'y′ and y′′y''y′′ back into the DE xy′′+y′=2xxy''+y'=2xxy′′+y′=2x to see if your solution is valid

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Case 2: Independent Variable x is Missing
If our DE is of the form F(y, y′, y′′)=0F\left(y,\ y',\ y''\right)=0F(y, y′, y′′)=0, then
we let v=y′v=y'v=y′ and v dvdy=y′′v\ \frac{dv}{dy}=y''v dydv​=y′′
solve the DE for the solution v=...v=...v=...
solve for the solution y=...y=...y=... using v=dydxv=\frac{dy}{dx}v=dxdy​
Example:
Solve the DE y′′=2yy′y''=2yy'y′′=2yy′

1. Let  v=y′v=y'v=y′ and v dvdy=y′′v\ \frac{dv}{dy}=y''v dydv​=y′′ to get v dvdy=2yvv\ \frac{dv}{dy}=2yvv dydv​=2yv

2. Now we solve this DE in terms of v and y:
dvdy=2y\frac{dv}{dy}=2ydydv​=2y

So we can solve this like a separable DE
1 dv=2ydy1\ dv=2ydy1 dv=2ydy
∫1 dv=∫2y dy\int_{ }^{ }1\ dv=\int_{ }^{ }2y\ dy∫​1 dv=∫​2y dy
v=y2+Cv=y^2+Cv=y2+C

3. Finally, since dydx=v\frac{dy}{dx}=vdxdy​=v:
dydx=y2+C\frac{dy}{dx}=y^2+Cdxdy​=y2+C

Solving this like a separable DE
1y2+Cdy=1 dx\frac{1}{y^2+C}dy=1\ dxy2+C1​dy=1 dx
1Carctan⁡(yC)=x+D\frac{1}{C}\arctan\left(\frac{y}{\sqrt{C}}\right)=x+DC1​arctan(C​y​)=x+D
arctan⁡(yC)=Cx+CD\arctan\left(\frac{y}{\sqrt{C}}\right)=Cx+CDarctan(C​y​)=Cx+CD
yC=tan⁡(Cx+CD)\frac{y}{\sqrt{C}}=\tan\left(Cx+CD\right)C​y​=tan(Cx+CD)
y=Ctan⁡(Cx+CD)\boxed{y=\sqrt{C}\tan\left(Cx+CD\right)}y=C​tan(Cx+CD)​

Wize University Calculus 2 Textbook > Differential Equations

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Second-Order Reducible DE

Case 1: Dependent Varible y is Missing

Case 2: Independent Variable x is Missing

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