Question

Solve the differential equation:

`y + d/dx (xy) = x (sinx + x)`

Answer 1

`y + d/dx (xy) = x (sinx + x)`

Using the product formula:

`y + (d(x))/dxy + x (dy)/dx = x (sinx + x)`

`y + y + x (dy)/dx = x(sin x + x)`

`x (dy)/dx + 2y = x(sin x + x)`

Diving both sides by x:

`x/x xx (dy)/dx + (2y)/x = (x(sin x + x))/x`

`(dy)/dx + (2y)/x = (sin x + x)`

Comparing with `dy/dx + Py = Q`

P = `2/x` and Q = (sin x + x)

Finding Integrating factor (I.F):

I.F = `e^(int p dx)`

= `e^(int 2/x dx)`

= `e^(2int1/x dx)`

= `e^(2 log |x|)`

= `e^(log x^2)`

= x²

Solution of differential equation is:

y × I.F = ∫ Q × IF dx

Putting values:

`y xx x^2 = ∫(sin x + x) x^2 dx`

`yx^2 = ∫ sin x × x^2 dx + ∫ x^3 dx`

`yx^2 = int sin x xx x^2 dx + x^4/4 + C`

`yx^2 = -x^2 cos x + 2 ∫ x cos x  dx + x^4/4 + C`

`yx^2 = int x^2 sin x  dx + x^4/4 + C`  .....(1)

Evaluating `int x^2 sin x  dx` separately;

`int x^2 sin x  dx = x^2 int sin x  dx - int((d(x^2))/dx int sin x  dx) dx`

= `-x^2 cos x - int 2x xx -cos x  dx`

= `-x^2 cos x + 2 int x cos x  dx + C`

Applying by parts again in `int x cos x`

= `-x^2 cos x + 2[x intcos x  dx - int(dx/dx int cos x  dx) dx] + C`

= `-x^2 cos x + 2[x sin x - int 1 xx sin x  dx] + C`

= `-x^2 cos x + 2 [x sin x - int sin x  dx] + C`

= −x² cos x + 2[x sin x − (−cos x)] + C

= −x² cos x + 2[x sin x + cos x] + C

Putting value of `int x^2 sin x  dx` in (1)

`yx^2 = int x^2 sin x  dx + x^4/4 + C`

`yx^2 = -x^2 cos x + 2[x sin x + cos x] + x^4/4 + C`

Which is the required solution.