What is the solution for (x-2)y^'-y=2(x-2)^3,y(3)=5 ?

The solution for (x-2)y^'-y=2(x-2)^3,y(3)=5 is y=x^3-6x^2+16x-16

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(x-2)y^'-y=2(x-2)^3,y(3)=5

(x - 2) y^{^{'}} - y = 2 (x - 2)^{3}, y (3) = 5

y = x^{3} - 6 x^{2} + 16 x - 16

Solution steps

(x - 2) y^{'} (x) - y = 2 (x - 2)^{3}

Solve linear ODE:

y = x^{3} - 6 x^{2} + 16 x - 16

y = x^{3} - 6 x^{2} + 16 x - 16

\int \frac{2 x ln ( x ^{2} + 1 )}{x ^{2} + 1} d x

x^{2} + 16, at x = 3

\int \frac{1}{3} x^{3} 9 - x^{2} d x

\frac{d}{d x} (2 x^{2} + 5)

y = cot^{2} (cos (x))

What is the solution for (x-2)y^'-y=2(x-2)^3,y(3)=5 ?
The solution for (x-2)y^'-y=2(x-2)^3,y(3)=5 is y=x^3-6x^2+16x-16