What is the solution for 4y^{''}-20y^'+25y=6-5x+50x^2 ?

The solution for 4y^{''}-20y^'+25y=6-5x+50x^2 is y=c_{1}e^{(5x)/2}+c_{2}e^{(5x)/2}x+2x^2+3x+2

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4y^{''}-20y^'+25y=6-5x+50x^2

4 y^{^{''}} - 20 y^{^{'}} + 25 y = 6 - 5 x + 50 x^{2}

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

Solution steps

4 y^{''} (x) - 20 y^{'} (x) + 25 y = 6 - 5 x + 50 x^{2}

Solve linear ODE:

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

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

y^{^{''}} - 4 y^{^{'}} = sin^{2} (x)

y^{^{''}} - 4 y^{^{'}} + 5 y = 5 x^{2} + 2 x

y^{^{''}} - y^{^{'}} + 4 = 3 x^{2} + 4 x

y^{^{''}} - y^{^{'}} + 4 y = 5 t^{2} + 9 t + 4

y^{^{'''}} + 3 y^{^{'}} + 2 = 0

What is the solution for 4y^{''}-20y^'+25y=6-5x+50x^2 ?
The solution for 4y^{''}-20y^'+25y=6-5x+50x^2 is y=c_{1}e^{(5x)/2}+c_{2}e^{(5x)/2}x+2x^2+3x+2