 # Solve for the general solution of the given special second-ordered differential equation xy''+x(y')^2-y'=0 postillan4 2020-11-09 Answered
Solve for the general solution of the given special second-ordered differential equation
$x{y}^{″}+x\left({y}^{\prime }{\right)}^{2}-{y}^{\prime }=0$
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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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