# Find the general solution of the given differential equation. x (dy/dx) + 4y = x^{3} - x

**Solution:**

Given,

Differential equation x (dy/dx) + 4y = x^{3} - x

Dividing both sides by x,

dy/dx + 4y/x = x^{2} - 1 --- (1)

Rewriting the given equation in first order differential form

(dy/dx) + Py = Q

Where, P = 4/x

Q = x^{2} - 1

Now, integrating factor,

I.F. = e^{∫p dx}

I.F. = e^{∫(4/x)} dx

= x^{4}

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

I.F × y = ∫Q × I.F. dx

x^{4} × y = ∫(x^{2} - 1)x^{4}. Dx

x^{4}(dy/dx) + 4x^{3}y = x^{6}−x^{4}

d(x^{4}y)/ dx = x^{6}−x^{4}

d(x^{4}y) = (x^{6}−x^{4}) dx

∫d(x^{4}y) = ∫(x^{6}−x^{4}) dx

x^{4}y = (1/7)x^{7}−(1/5)x^{5 }+ C

Dividing both sides by x^{4},

y = (1/7)x^{3}−(1/5)x^{ }+ C/x^{4}

Therefore, the general solution is y = (1/7)x^{3}−(1/5)x^{ }+ C/x^{4}

## Find the general solution of the given differential equation. x (dy/dx) + 4y = x^{3} - x

**Summary:**

The general solution of the given differential equation x (dy/dx) + 4y = x^{3} - x is y = (1/7)x^{3}−(1/5)x^{ }+ C/x^{4}

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