ASSIGNMENT 02(MTH401) Differential Equation

Question 1:                                                                                                                             

Solve the initial value problem.

4 dy / dx - 2y = xy⁵,     y(1) = 3

4 dy / dx - 2y = xy⁵ --------------------------------(1)

P(x) = -2   , q(x) = x  ,   n=5

Divide the equation (1) with y⁵ 

4y^-5 dy/dx -2y^-4 =x

Let suppose

w = y^-4

dw / dx = -4 y^-5 dy / dx ------------------------------------(2)

as we know from the original equation:

dy / dx = (xy5 + 2y) /4

putting the value of dy/dx in eq (2)

dw / dx = - y^-5 (xy⁵ +2y)

dw / dx = - x - 2y^-4

dw / dx = - x - 2w

dw / dx + 2w = -x

by taking integral factor

= e ^{f (p(x)dx)}

P(x) = 2

u(x) = e ^{f 2dx} = e^2x

Multiply the whole equation with integrating factor

dw / dx e^2x + 2w e^2x = -x e^2x

f d / dx (w e^2x )= f -x e^2x

we^2x = f -x e^2x

we^2x = - (x e^2x/2 - f  e^2x/2dx)

we^2x = - (x e^2x/2 - e^2x/4 + c)

w  = - x /2 + 1/4 - c e^-2x

y^-4 = - x /2 + 1/4 - c e^-2x

(3)^-4 = - 1 /2 + 1/4 - c e^-2

1/81 = - 1 / 4 - c e^-2

C =85e² / 204

y^-4 = - x /2 + 1/4 -  85e² / 204 e^-2x

y = ( - x /2 + 1/4 -  85e² / 204 e^-2x ) ^{-1 / 4}