Elementary Differential Equations by Dela Fuente, Feliciano, and Uy Chapter 9 Problem 6 — Special Second-Ordered Differential Equations

Solve the following differential equation

yy′′−(y′)2+y′=0yy''-\left(y'\right)^2+y'=0

Solutions:

Basically, We need to make the orders of each term to 1. To be able to further break down the equation.

so we let P=y′             Pdpdy=y′′so\:we\:let\:P=y' \\ \:\:\:\:\:\:\:\:\:\:\:\:\:P\frac{dp}{dy}=y''

Substituting to the equation, we get

yPdPdy−P2+P=0yP\frac{dP}{dy}-P^2+P=0

Removing the variables y and P from the 1st term we get

dPdx−Py+1y=0      ,the equation has become a FOLDE  dPdy−Py=−1y           T(y)=−1y,     Q(y)=−1y     ϕ=e∫ −1ydy=y−1         Pϕ=∫ϕ Q(y)dy + C1            Py−1=∫ y−1(−y−1)dy+C1Py−1=∫ −y−2dy+C1 Py=1y+C1    P=1+yC1dydx=1+yC1\frac{dP}{dx}-\frac{P}{y}+\frac{1}{y}=0\:\:\:\:\:\: , the \:equation\:has\:become\:a \:FOLDE\\ \;\\ \frac{dP}{dy}-\frac{P}{y}=-\frac{1}{y}\:\:\:\:\:\:\:\:\:\:\:T\left(y\right)=-\frac{1}{y},\:\:\:\:\:Q\left(y\right)=-\frac{1}{y}\\ \:\\\:\:\:\: \phi =e^{\int \:-\frac{1}{y}dy}\\ =y^{-1} \\\:\:\:\:\:\:\:\:\:P\phi=\int\phi\:Q(y)dy\:+\:C_{1} \\\:\:\:\:\:\:\:\:\:\:\:\:Py^{-1}=\int \:y^{-1}\left(-y^{-1}\right)dy+C_{1} \\ Py^{-1}=\int \:-y^{-2}dy+C_{1} \\ \:\\ \frac{P}{y}=\frac{1}{y}+C_{1} \\\: \:\\\:\: P=1+yC_{1} \\ \frac{dy}{dx}=1+yC_{1}

By means of Separation of Variables

dy1+yC1= dx∫ dy1+yC1=∫ dx        let u=1+yC1      du=C1dyduC1=dy   1C1∫ duu=∫ dx           1C1ln u=x+C2\\ \frac{dy}{1+yC_{1}}=\:dx \\ \int \:\frac{dy}{1+yC_{1}}=\int \:dx \\\:\:\:\:\:\:\:\: let\:u=1+yC_{1} \\\:\:\:\:\:\: du=C_{1}dy\\ \frac{du}{C_{1}}=dy \\\:\:\: \frac{1}{C_{1}}\int \:\frac{du}{u}=\int \:dx \\\:\:\:\:\:\:\:\:\:\:\: \frac{1}{C_{1}}ln\:u=x+C_{2}

We get

ln∣1+yC1∣=C1x+C2ln\left|1+yC_{1}\right|=C_{1}x+C_{2}
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