Answer to Question #197056 in Differential Equations for Rabia

Question

Determine the Integrating factor and identify the solution of general/initial value problem

a) (𝑥2 +1)𝑦′ +2𝑥𝑦=4𝑥2 ; 𝑦(0)=0

1+𝑥2

b)𝑦′+ 𝑦 =3𝑥2 𝑥𝑙𝑛𝑥 ln 𝑥

c) (𝑥2 + 1) 𝑦′ + 4𝑥𝑦 = 𝑥 ; 𝑦(2) = 1

Answer 1

(a) Given equation-

$(x^2+1)y'+2xy=4x^2; y(0)=0$

The simplifying expression is-

$\dfrac{dy}{dx}+\dfrac{2xy}{x^2+1}=\dfrac{4x^2}{x^2+1}$

Integrating factor-

$I.F.=e^{\int Pdx}$

  $=e^{\frac{2x}{1+x^2}}dx \\=e^{ln(1+x^2)} \\ =1+x^2$

Solution is-

$y\times (1+x^2)=\int (1+x^2)(\dfrac{4x^2}{1+x^2})dx \\[9pt] \Rightarrow y(1+x^2)=\dfrac{4x^3}{3}+C$

Putting Y(0)=0

$0=0+C\Rightarrow C=0$

So Solution is-

$(1+x^2)y=\dfrac{4x^3}{3}$

(b)Given equation-

  $y'+y=3x^2 \\[9pt] \dfrac{dy}{dx}+y=3x^2$

Integrating factor-

I$.F.=e^{1.dx}=e^x$

Solution is-

$y\times e^x=\int e^x 3x^2dx \\[9pt] ye^x=3x^2e^x-6xe^x+6e^x+C \\[9pt] y=3x^2-6x+6+ce^{-x}$

(C) Given equation is-

$(x^2+1)y'+4xy=x;y(2)=1$

The simplified expression is-

$\dfrac{dy}{dx}+\dfrac{4xy}{x^2+1}=\dfrac{x}{x^2+1}$

Integrating factor-

$I.F.=e^{\frac{4x}{1+x^2}dx}=e^{2log(1+x^2)}=e^{log(1+x^2)^2}=(1+x^2)^2$

Solution is-

$y\times x^2=\int (1+x^2)^2 \dfrac{x}{x^2+1}dx \\[9pt] \Rightarrow yx^2=\int(x+x^3)dx\\[9pt] \Rightarrow yx^2=\dfrac{x^2}{2}+\dfrac{x^4}{4}+c \\[9pt] \Rightarrow y=\dfrac{1}{2}+\dfrac{x^2}{4}+\dfrac{c}{x^2}$

Also, y(2)=1

$\Rightarrow 1=\dfrac{1}{2}+\dfrac{4}{4}+\dfrac{c}{4}\\[9pt] \dfrac{c}{4}=-\dfrac{1}{2}\\[9pt]c=-2$

So solution is-

$y=\dfrac{1}{2}+\dfrac{x^2}{4}-\dfrac{2}{x^2}$