# Find the general solution of the given differential equation. x^{2}y' + xy = 8. y(x) = give the largest interval over which the general solution is defined.

**Solution:**

Given, x^{2}y' + xy = 8

⇒ x^{2} .dy/dx + x . y = 8

Divide both the sides by x^{2}

dy / dx + y / x = 8 / x^{2}

It is of the form dy / dx + P(x)y = Q(x), which is a linear differential equation where, P(x) = 1 / x and Q(x) = 4 / x^{2}.

Integrating factor (I. F.) = e^{ ∫P(x). dx}

I.F. = e ^{∫1/x. dx}

= e^{logex}

= x

Solution of Linear differential equation is

(I.F.) × y = ∫ (I.F.) Q(x) dx

x y = ∫ x. (8 / x^{2}) dx

xy = 8 ∫ (1 / x) . dx

xy = 8 log_{e}x + C

This general solution is defined as ∀ x ϵ R^{+} because if x = 0 or x = -ve, log_{e}x does not exist.

## Find the general solution of the given differential equation. x^{2}y' + xy = 8. y(x) = give the largest interval over which the general solution is defined.

**Summary:**

The general solution of the given differential equation. x^{2}y' + xy = 8 is xy = 8 log_{e}x + C.

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