# Find the general solution of the given differential equation. x (dy/dx) + 3y = x^{3} - x

**Solution:**

Given,

Differential equation x (dy/dx) + 3y = x^{3} - x

Dividing both sides by x,

dy/dx + 3y/x = x^{2} - 1 --- (1)

Rewriting the given equation in first order differential form

(dy/dx) + Py = Q

Where, P = 3/x

Q = x^{2 }- 1

Now, integrating factor,

I.F. = e^{∫p dx}

I.F. = e^{∫(6/x) dx}

= x^{6}

On multiplying equation (1) by I.F., we get,

I.F × y = ∫Q × I.F. dx

x^{6} × y = ∫(x^{2} - 1)x^{6}. dx

d(x^{6}y) = ∫(x^{8} - x^{6}).dx

d(x^{6}y) = (x^{8} - x^{6}) dx

∫d(x^{6}y) = ∫(x^{8} - x^{6}) dx

x^{6}y = (1/9)x^{9} - (1/7)x^{7} + C

Dividing both sides by x^{6},

y = (1/9)x^{3} - (1/7)x + c/x^{6}

Therefore, the general solution is y = (1/9)x^{3} - (1/7)x + c/x^{6}.

## Find the general solution of the given differential equation. x (dy/dx) + 3y = x^{3} - x

**Summary:**

The general solution of the given differential equation x (dy/dx) + 3y = x^{3} - x is y = (1/9)x^{3} - (1/7)x + c/x^{6}.

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