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