Question

solve x dy/dx-y=(x-1)ex​

Answer 1

Please see the attachment

Answer 2

The solution of xdy/dx - y = (x - 1)e^x is

y = e^x + Cx

• We have ,

xdy/dx - y = (x - 1)e^x

Dividing the above equation by x, we get

dy/dx - y(1/x) = (x - 1)/x (e^x)

• Now this differential equation is in the form

dy/dx - y P(x) = Q(x)

where P(x) = -1/x and Q(x) = (x - 1)/x (e^x)

• Now we will find the integrating factor (I.F)

I.F = e^[∫ P(x)dx]

= e^[∫( -1/x) dx]

= e^(-logx)

= e^(log(1/x))

I.F = 1/x - (1)

• The solution is given by

y (I.F) = ∫ Q(x) (I.F) dx

Now, putting the value of I.F from (1) in the above equation ,we get

y(1/x) = ∫(1/x) (x - 1)/x (e^x) dx

y(1/x) = ∫ e^x (x - 1)/x^2 dx

y(1/x) = ∫ e^x (1/x - 1/x^2)dx - (2)

• Now RHS is in the form ∫e^x [ f(x) + f'(x) ] dx

where f(x)= 1/x and f'(x) = -1/x^2

it's integration is given by

∫e^x [ f(x) + f'(x) ] dx = (e^x) f(x) + C

therefore,

∫ e^x (1/x - 1/x^2)dx = e^x (1/x) + C - (3)

• Equating (2) and (3) we get,

y(1/x) = e^x (1/x) + C

therefore, y = e^x + Cx