Question

solve x^2d^2y/dx^2 -3xdy/dx +4y =2x^2​

Answer 1

Solve the Differential equation :-

$\sf \: {x}^{2} \dfrac{ {d}^{2}y }{ {dx}^{2} } - 3x\dfrac{dy}{dx} + 4y = {2x}^{2} - - - (1)$

Now,

$\sf \: Put \: x = {e}^{z} \: \implies \: z = log(x)$

Now,

$\sf \: As \: D = x\dfrac{d}{dx} \: where \: D \: = \: \dfrac{d}{dz} - - - (2)$

and

$\sf \: {x}^{2} \dfrac{ {d}^{2} }{ {dx}^{2} } = D( D - 1) - - - (3)$

Substituting all these values in equation (1), we get

$\sf \: \{ D(D - 1) - 3D + 4 \}y = 2 {e}^{2z}$

$\sf \: \{ {D}^{2} - D- 3D + 4 \}y = 2 {e}^{2z}$

$\sf \: \{ {D}^{2} - 4D + 4 \}y = 2 {e}^{2z}$

$\sf \: {(D - 2)}^{2} y = 2 {e}^{2z}$

Therefore, Auxiliary equation is

$\sf \: {(D - 2)}^{2} = 0$

$\sf \: \therefore \: D \: = 2 \: or \: 2$

So,

Complementary function,

$\sf \: C.F. = (c_1+c_2z) {e}^{2z}$

$\sf \: C.F. = (c_1+c_2 log(x) ) {x}^{2}$

Now,

Particular integral is,

$\sf \: P. I. = \dfrac{1}{ {(D - 2)}^{2} } {2e}^{2z}$

$\sf \: P. I. = 2z\dfrac{1}{2(D - 2)} {e}^{2z} \: \: \: \: (case \: of \: failure)$

$\sf \: P. I. = {z}^{2} {e}^{2z}$

$\sf \: P. I. = {x}^{2} {(logx)}^{2}$

Hence,

Complete Solution, y = C. F. + P. I.

$\therefore \: \bf \: y \: = \: (c_1+c_2 log(x) ) {x}^{2} + {x}^{2} {(logx)}^{2}$

Answer 2

Answer:

Common Solution Y= CF+PIY

Step-by-step explanation: