Solve the differential equation : (1 + x^2)\frac{dy}{dx} +2xy -4x^2 =0, subject to the initial condition y(0) = 0

 

 

 

 
 
 
 
 

Answers (1)

Given : (1+x^2)\frac{\mathrm{d} y}{\mathrm{d} x}+2xy-4x^2=0

\Rightarrow \frac{\mathrm{d} y}{\mathrm{d} x}+\frac{2xy}{1+x^2}-\frac{4x ^2}{1+x^2}=0

\Rightarrow \frac{\mathrm{d} y}{\mathrm{d} x}+\frac{2xy}{1+x^2}=\frac{4x ^2}{1+x^2}

\frac{\mathrm{d} y}{\mathrm{d} x}+Py=Q

Here :  P=\frac{2x}{1+x^2},       Q=\frac{4x^2}{1+x^2}

I.F =e^{\int\! P \: dx}=e^{\int\! \frac{2x}{1+x^2}}=e^{\ln (1+x^2)}=(1+x^2)

Solution is : 

y(I.F)=\int\! Q\cdot I.F\: dx +c

y(1+x^2)=\int\!\frac{4x^2}{1+x^2}\cdot (1+x^2) dx+c

y(1+x^2)=4\int\!x^2 dx+c

y(1+x^2)=4\frac{x^3}{3}+c\: \: \: \: \: \: -(1)

Applying condition in (1),

x=0,y=0

0(1+0)=4\left ( \frac{0}{3} \right )+c\Rightarrow c=0

Putting c value in (1)

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

or  y=\frac{4x^3}{3(1+x^2)}

Preparation Products

Knockout KCET 2021

An exhaustive E-learning program for the complete preparation of KCET exam..

₹ 4999/- ₹ 2999/-
Buy Now
Knockout KCET JEE Main 2021

It is an exhaustive preparation module made exclusively for cracking JEE & KCET.

₹ 27999/- ₹ 16999/-
Buy Now
Knockout NEET Sept 2020

An exhaustive E-learning program for the complete preparation of NEET..

₹ 15999/- ₹ 6999/-
Buy Now
Rank Booster NEET 2020

This course will help student to be better prepared and study in the right direction for NEET..

₹ 9999/- ₹ 4999/-
Buy Now
Knockout JEE Main Sept 2020

An exhaustive E-learning program for the complete preparation of JEE Main..

₹ 12999/- ₹ 6999/-
Buy Now
Exams
Articles
Questions