What is the solution for x^2y^{''}-7xy^'+16y=0 ?

The solution for x^2y^{''}-7xy^'+16y=0 is y=c_{1}x^4+c_{2}ln(x)x^4

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x^2y^{''}-7xy^'+16y=0

x^{2} y^{^{''}} - 7 x y^{^{'}} + 16 y = 0

y = c_{1} x^{4} + c_{2} ln (x) x^{4}

Solution steps

x^{2} y^{''} (x) - 7 x y^{'} (x) + 16 y = 0

Solve linear ODE:

y = c_{1} x^{4} + c_{2} ln (x) x^{4}

y = c_{1} x^{4} + c_{2} ln (x) x^{4}

a re a y = - 3 x^{3} + x^{2} + 10 x, x \geq 0

y^{^{''}} - 2 y^{^{'}} - 3 y = x e^{x}

d er i v a t i v e e^{\frac{4}{5} x}

\frac{d}{d x} (x + 1 - 2)

\frac{d}{d x} (\frac{e ^{4 x^{2 + 1}}}{8})

What is the solution for x^2y^{''}-7xy^'+16y=0 ?
The solution for x^2y^{''}-7xy^'+16y=0 is y=c_{1}x^4+c_{2}ln(x)x^4