Solve for the general solution of the given special second-ordered differential equation xy''+x(y')^2-y'=0

Solve for the general solution of the given special second-ordered differential equation xy''+x(y')^2-y'=0

Question
Solve for the general solution of the given special second-ordered differential equation
\(xy''+x(y')^2-y'=0\)

Answers (1)

2020-11-10
The second order differential equation is given as:
\(xy''+x(y')^2-y'=0\)
isolated y''
\(y''=-\frac{x(y')^2-y'}{x}\)
Substitute y'=v and v is the function of x:
\(y''=-\frac{x(y')^2-y'}{x}\)
\(y'=v\)
\(v'=-\frac{x(v)^2-v}{x}\Rightarrow\frac{dv}{dx}=-\frac{x(v)^2-v}{x}\)
Now, the equation is of thr form:
A first order Bernoulli IDE has the form of
\(y'+p(x)y=q(x)y^n\)
\(v'-\frac{1}{x}v=-v^2\)
The general solution is obtained by substituting \(u=v^(1-n)\)
solving \frac{1}{1-n}u'+p(x)u=q(x)\)
put \(u=v^{-1}\)
\(-u'-\frac{u}{x}=-1\Rightarrow u'+\frac{u}{x}=1\)
Rewrite the first irderliner ODE and solve the equation:
\(u'+\frac{u}{x}=1\)
Rewrite in the form of a first order linear ODE.
\(u'(x)+p(x)\cdot u=q(x)\)
\(p(x)=\frac{1}{x},\ q(x)=1\)
\(I.F.=e^{\int\frac{1}{x}dx}=x\)
\(u(x)\cdot x=\int1\cdot u(x)dx=\int xdx=\frac{x^2}{2}+c_1\)
\(u(x)=\frac{x}{2}+\frac{c_1}{x}=\frac{x^2+2c_1}{2x}\) Let \(2c_1=c\)
\(v(x)=\frac{1}{u(x)}=\frac{2x}{x^2+2c_1}\)
Substitute back the values of \(v(x)=y'\) and obtain the solution:
Substitute back \(v=y'\)
\(v(x)=y'(x)=\frac{2x}{x^2+c}\)
\(\frac{dy}{dx}=\frac{2x}{x^2+c}\Rightarrow dy=\frac{2x}{x^2+c}dx\)
integrating both sides:
\(\int dy=\int\frac{2x}{x^2+c}dx\)
\(y+c_2=\log(x^2+c)\)
\(y=\log(x^2+c)-c_2\)
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