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

We will be using the First Order differential equation to solve this.

## Answer: The general solution of the given differential equation x(dy/dx) + 4y = x^{3} - x is (1/7)X^{3}- (1/5)x + (c /x^{4})

Let's solve this question step by step.

**Explanation: **

Given:

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

Dividing both sides by x,

(dy/dx) + 4y/x = x^{2} − 1....(eq 1)

To solve the equation (dy/dx) + 4y/x = x^{2} − 1 we need to rewrite the given differential equation in First Order DE 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 (eq 1) by this 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

y = (1/7)x^{3}−(1/5)x^{ }+ C/x^{4}