What is the solution for y^{''}-10y^'+25y=30x+3 ?

The solution for y^{''}-10y^'+25y=30x+3 is y=c_{1}e^{5x}+c_{2}xe^{5x}+(6x)/5+3/5

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y^{''}-10y^'+25y=30x+3

y^{^{''}} - 10 y^{^{'}} + 25 y = 30 x + 3

y = c_{1} e^{5 x} + c_{2} x e^{5 x} + \frac{6 x}{5} + \frac{3}{5}

Solution steps

y^{''} (x) - 10 y^{'} (x) + 25 y = 30 x + 3

Solve linear ODE:

y = c_{1} e^{5 x} + c_{2} x e^{5 x} + \frac{6 x}{5} + \frac{3}{5}

y = c_{1} e^{5 x} + c_{2} x e^{5 x} + \frac{6 x}{5} + \frac{3}{5}

f (x) = \frac{3}{x + 5}, at x = 2

\frac{\partial}{\partial y} (- x - sin (y^{2}))

\int x^{3} (x - 1)^{- 4} d x

\frac{s ^{4}}{s}

f (x) = 3 x^{2} + 3 x + 4

What is the solution for y^{''}-10y^'+25y=30x+3 ?
The solution for y^{''}-10y^'+25y=30x+3 is y=c_{1}e^{5x}+c_{2}xe^{5x}+(6x)/5+3/5