# Solve the Differential Equation: ( 1 + X 2 ) D Y D X + 2 X Y − 4 X 2 = 0 , Subject to the Initial Condition Y(0) = 0. - Mathematics

Sum

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

#### Solution

The given differential equation can be written as:

dy/dx + (2x)/(1+x^2)y = (4x^2)/(1+x^2)  ...(1)

This is a linear differential equation of the form dy/dx + Py = Q

P = (2x)/(1+x^2) and Q = (4x)/(1+x^2)

"I.F" = e^(intPdx) = e^(int(2x)/(1+x^2)dx) = e^log(1+x^2) = 1+x^2

Multipying both sides of  (1) by I.F = (1+x^2), we get

(1+x^2)dy/dx + 2xy = 4x^2

Integrating both sides with respect to x, we get

y (1+x^2) = int4x^2dx +"C"

y (1+x^2) = (4x^3)/(3) + "C" ...(2)

Given y = 0, when x =0

Substituting x = 0 and y = 0 in (1), we get

0 = 0 + C ⇒ C = 0

Substituting C = 0 in (2), we get y = (4x^3)/(3(1+x^2), which is the required solution.

Concept: Methods of Solving First Order, First Degree Differential Equations - Linear Differential Equations
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