What is the solution for (x^2-9)y^'+6y=(x+3)^2 ?

The solution for (x^2-9)y^'+6y=(x+3)^2 is y=(x(x+3))/(x-3)+(c_{1}(x+3))/(x-3)

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(x^2-9)y^'+6y=(x+3)^2

(x^{2} - 9) y^{^{'}} + 6 y = (x + 3)^{2}

y = \frac{x ( x + 3 )}{x - 3} + \frac{c _{1} ( x + 3 )}{x - 3}

Solution steps

(x^{2} - 9) y^{'} (x) + 6 y = (x + 3)^{2}

Solve linear ODE:

y = \frac{x ( x + 3 )}{x - 3} + \frac{c _{1} ( x + 3 )}{x - 3}

y = \frac{x ( x + 3 )}{x - 3} + \frac{c _{1} ( x + 3 )}{x - 3}

x \cdot sin (x^{2})

x^{2} + x y - y^{2} = 19, (4, 1)

x \to 0 + lim (\frac{cos ( x ) - 1}{x})

\int_{- 3}^{3} 1 + (x)^{2} d x

x^{2} - 4 x + 8

What is the solution for (x^2-9)y^'+6y=(x+3)^2 ?
The solution for (x^2-9)y^'+6y=(x+3)^2 is y=(x(x+3))/(x-3)+(c_{1}(x+3))/(x-3)