Solve the following differential equation: `(x^2-1)dy/dx+2xy=2/(x^2-1)`

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#### Solution

we have:

`dy/dx+2x/(x^2-1)y=2/(x^2-1)^2`

This is a linear differential equation of the form `dy/dx+Py=Q` ,where `P=(2x)/(x^2-1) and Q=2/(x^2-1)^2`

`I.f=e^(intPdx)=e^(int(2x)/(x^2-1)dx)=e^(log(x^2-1))=(x^2-1)`

Hence, the solution of the differential equation is given by

`y.(x^2−1)=∫2/(x^2−1)^2×x(x2−1) dx + C`

`y.(x^2−1)=∫2/(x^2−1) dx + C`

We know that,

`∫dx/(x^2−a^2)=1/(2a)log∣(x−1)/(x+1)∣`

`y.(x^2−1)=2xx1/2log∣(x−1)/(x+1)∣+C`

Concept: Solutions of Linear Differential Equation

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