Solve the dierential equation
$y 002 y 0 + y = 2 e x l n x, x > 0. 2$

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Solution

The solution of the D.E. is given by $y = y c + y p$ where $y c$ is the solution to $y 002 y 0 + y = 0$ and yp is a particular solution. We have $r 22 r + 1 = (r 1) 2 = 0$ or $r = 1.$ Thus, $y c = c 1 e x + c 2 x e x .$
$y p$ is given by $y p = e x u 1 + x e x u 2$ where
$e x u 01 + x e x u 02 = 0$
$e x u 01 + (x + 1) e x u 02 = 2 e x l n x$
we have
$u 01 = 0 x e x$
$e x x e x = - 2 x e 2 x l n x e 2 x = 2 x l n x o r, u 1 = Z 2 x l n x d x = Z l n x d (x 2) = x 2 l n x + Z x 21 x d x$ or
$u 1 = x 2 l n x + x 22$
$u 02 = e x 0$
$e x x e x = 2 e 2 x l n x e 2 x = 2 l n x o r u 2 = 2 l n x d x = 2 x l n x 21 x x d x = 2 x l n x 2 x .$ $T h e r e f o r e, y p = e x (x 2 l n x + x 22) + x e a x (2 x l n x 2 x) 3$
and
$y = e x [c 1 + c 2 x x 2 l n x + x 22 + 2 x 2 l n x 2 x 2]$
or
$y = e x [c 1 + c 2 x + x 2 l n x 32 x 2]$

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Solve the equation by substitution method

0.2x+0.3y=1.3

0.4x+0.5y=2.3

(1) Draw the graph of y = x² and y = 2x + 3 and hence solve the equation x² − 2x − 3 = 0.

(2) Draw the graph of y = 2x²and y = 3 − x and hence solve the equation 2x² + x − 3 = 0.

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