Question

(x^2-x-2)y’+3xy=x^2-4x+4

Answer 1

The given D.E is a lin ear 1st order differential equation:

$$(x^2-x-2)\frac{dy}{dx}+3xy =( x^2-4x+4)$$

Now this can be rewritten as

$$\frac{dy}{dx}+\frac{3x}{(x^2-x-2)}*y=\frac{(x^2-4x+4)}{(x^2-x-2)}$$

$or, \frac{dy}{dx}+\frac{3x}{(x+1)(x-2)}*y=\frac{(x-2)^2}{(x+1)(x-2)}\\or, \frac{dy}{dx}+\frac{3x}{(x+1)(x-2)}*y=\frac{(x-2)}{(x+1)}$

Calculating INTEGRATING FACTOR (I.F)

$I.F= e^{\int {\frac{3x}{(x+1)(x-2)}}dx}$

$Now \int {\frac{3x}{(x+1)(x-2)}}dx = \int \{\frac{1}{x+1}+\frac{2}{x-2}\}dx$

$=\int \frac{1}{x+1}dx + 2\int \frac{1}{x+2}dx$

$= log|x+1|+2log|x+2|$

$= log\{(x+1)(x-2)^2\}$

$\therefore I.F = e^{log\{(x+1)(x-2)^2\}}=(x+1)(x-2)^2$

The solution to the given D.E is

$$y*I.F=\int I.F*\frac{(x-2)}{(x+1)}dx + C$$

$\therefore y* (x+1)(x-2)^2= \int \{(x+1)(x-2)^2\frac{(x-2)}{(x+1)}\}dx +C\\or, y* (x+1)(x-2)^2= \int (x-2)^3dx+C$

$or,y* (x+1)(x-2)^2= \frac{(x-2)^4}{4}+C$

The required solution is

$$y* (x+1)(x-2)^2= \frac{(x-2)^4}{4}+C$$

ANSWER