What is the solution for y^'+1/(x+7)y=x^{-2},y(1)=4 ?

The solution for y^'+1/(x+7)y=x^{-2},y(1)=4 is y=(ln(x))/(x+7)-7/(x(x+7))+(39)/(x+7)

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y^'+1/(x+7)y=x^{-2},y(1)=4

y^{^{'}} + \frac{1}{x + 7} y = x^{- 2}, y (1) = 4

y = \frac{ln ( x )}{x + 7} - \frac{7}{x ( x + 7 )} + \frac{39}{x + 7}

Solution steps

y^{'} (x) + \frac{1}{x + 7} y = x^{- 2}

Solve linear ODE:

y = \frac{ln ( x )}{x + 7} - \frac{7}{x ( x + 7 )} + \frac{39}{x + 7}

y = \frac{ln ( x )}{x + 7} - \frac{7}{x ( x + 7 )} + \frac{39}{x + 7}

\frac{\partial}{\partial x} (x cos (x) cos (y))

x \to 2 lim (\frac{3 x ^{2} - 5 x - 2}{2 x - 1})

y^{2} d x = (2 x y + x^{2}) d y

\int 4 e^{- 5 x} d x

What is the solution for y^'+1/(x+7)y=x^{-2},y(1)=4 ?
The solution for y^'+1/(x+7)y=x^{-2},y(1)=4 is y=(ln(x))/(x+7)-7/(x(x+7))+(39)/(x+7)