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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.

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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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