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Wize University Calculus 1 Textbook > Differential Equations

Second-Order Reducible DEs

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Second-Order Reducible DE

These have the form F(x, y, y′, y′′)=0F\left(x,\ y,\ y',\ y''\right)=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)=0, then
  1. we let v=y′v=y' and v′=y′′v'=y''
  2. solve the DE for the solution v=...v=...
  3. solve for the solution y=...y=... using v=dydxv=\frac{dy}{dx}

Example:
Solve the DE xy′′+y′=2xxy''+y'=2x
1. Let v=y′v=y' and v′=y′′v'=y'' to get xv′+v=2xxv'+v=2x

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

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

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

*You can always do a check by substituting y′y' and y′′y'' back into the DE xy′′+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)=0, then
  1. we let v=y′v=y' and v dvdy=y′′v\ \frac{dv}{dy}=y''
  2. solve the DE for the solution v=...v=...
  3. solve for the solution y=...y=... using v=dydxv=\frac{dy}{dx}
Example:
Solve the DE y′′=2yy′y''=2yy'
1. Let v=y′v=y' and v dvdy=y′′v\ \frac{dv}{dy}=y'' to get v dvdy=2yvv\ \frac{dv}{dy}=2yv

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

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

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

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

Practice: Second Order DE

Solve the DE x2y′′=3(y′)2x^2y''=3\left(y'\right)^2

Practice: Second Order DE

Solve the DE 2yy′′+(y′)2=02yy''+\left(y'\right)^2=0