Answer to Question #151302 in Differential Equations for md Ahasan

Question

(x^2+1)d^2y/dx^2-(2x)dx/dy+2y=6(x^2+1)^2 solve it ODE

Answer 1

$(x^2+1)y''-2xy'+2y=6(x^2+1)^2$

Set $y=vx$

$\implies y'=v+xv'$ ,$y''=2v'+v''$ .

Put this into the equation.

$(x^2+1)(2v'+xv'')-2x(v+xv')+2xv=6(x^2+1)^2\\ x(x^2+1)v''+2x^2v'+2v'-2xv-2x^2v'+2xv=6(x^2+1)^2\\ x(x^2+1)v''+2v'=6(x^2+1)^2\\$

Divide through by $x(x^2+1)$ .

$v''+ \frac{2}{x(x^2+1)}v'=\frac{6(x^2+1)}{x}\\$

Set $u=v',u'=v''$ .

$u'+ \frac{2}{x(x^2+1)}u=\frac{6(x^2+1)}{x}\\$

We will be needing an Integrating factor.

$I.F=e^{\int\frac{2}{x(x^2+1)}dx}=e^{\int\frac{2}{x}dx-{\int\frac{2}{x^2+1}dx}}\\ =e^{2\ln x-\ln x^2+1}=\frac{e^{\ln x}}{e^{\ln x^2+1}}=\frac{x^2}{x^2+1}$

Multiply through by the $I.F$

$\frac{x^2}{x^2+1}\left(u'+ \frac{2}{x(x^2+1)}u\right)=\frac{x^2}{x^2+1}\left(\frac{6(x^2+1)}{x}\\\right)\\ \frac{d(u\frac{x^2}{x^2+1})}{dx}=6x$

Integrate both sides

$u\frac{x^2}{x^2+1}=\int 6xdx\\ u\frac{x^2}{x^2+1}=3x^2+C\\ u=3(x^2+1)+\frac{C(x^2+1)}{x^2}\\ \text{But, } u=v'. \text{So,}$

$v'=3(x^2+1)+\frac{C(x^2+1)}{x^2}\\ \text{Integrate both sides}\\ v=x^3+3x+\frac{C(x^2-1)}{x}+K\\ \text{But, } v=\frac{y}{x}. \text{ So, }\\ y=x^4+3x^2+C(x^2-1)+Kx\\ y=x^4+(3+C)x^2+Kx-C$