Sum

Solve the following differential equation.

`x dy/dx + 2y = x^2 log x`

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#### Solution

`x dy/dx + 2y = x^2 log x`

Dividing throughout by x, we get

`dy/dx + 2/x y = x log x `

The given equation is of the form

`dy/dx + py = Q`

where, P =`2/x and Q = x log x `

∴ I.F. =`e ^int pdx = e ^2int1/xdx = e^2log|x| = e log |x^2| = x^2 `

∴ Solution of the given equation is

`y (I.F.) = int Q (I.F.) dx + c`

∴ `yx^2 = int (x log x )x^2 dx + c`

= `int x^3 log x dx +c`

= `log x int x^3 dx - int (d/dx log x int x^3 dx) dx+c`

= `x^4/4 log x - int 1/x (x^4/4) dx +c`

= `x^4/ 4 logx - 1/4 int x^3 dx +c`

∴ `yx^2 = x^4/4 logx - X^4/16 +c`

Concept: Differential Equations

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