What is the solution for 3y^{''}-13y^'+4y=8x+30 ?

The solution for 3y^{''}-13y^'+4y=8x+30 is y=c_{1}e^{4x}+c_{2}e^{x/3}+2x+14

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3y^{''}-13y^'+4y=8x+30

3 y^{^{''}} - 13 y^{^{'}} + 4 y = 8 x + 30

y = c_{1} e^{4 x} + c_{2} e^{\frac{x}{3}} + 2 x + 14

Solution steps

3 y^{''} (x) - 13 y^{'} (x) + 4 y = 8 x + 30

Solve linear ODE:

y = c_{1} e^{4 x} + c_{2} e^{\frac{x}{3}} + 2 x + 14

y = c_{1} e^{4 x} + c_{2} e^{\frac{x}{3}} + 2 x + 14

x sin (- 2 x) π

8 x^{2}

\int 5 x^{2} \cdot e^{5 x} d x

\frac{1}{2} (2 t + 1)^{- \frac{1}{2}}

a x^{2} e^{2 x}

What is the solution for 3y^{''}-13y^'+4y=8x+30 ?
The solution for 3y^{''}-13y^'+4y=8x+30 is y=c_{1}e^{4x}+c_{2}e^{x/3}+2x+14