What is the solution for y^{''}+1.5y^'-y=12x^2+6x^3-x^4 ?

The solution for y^{''}+1.5y^'-y=12x^2+6x^3-x^4 is y=c_{1}e^{x/2}+c_{2}e^{-2x}+x^4

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y^{''}+1.5y^'-y=12x^2+6x^3-x^4

y^{^{''}} + 1.5 y^{^{'}} - y = 12 x^{2} + 6 x^{3} - x^{4}

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

Solution steps

y^{''} (x) + 1.5 y^{'} (x) - y = 12 x^{2} + 6 x^{3} - x^{4}

Solve linear ODE:

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

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

\frac{d}{d x} (e^{m t} \cdot sin (n t))

f (x) = 3 cos (2 ln (x))

x \frac{d y}{d x} + y = x

y^{^{'}} + 6 y = e^{4 t}, y (0) = 2

\int \frac{x ^{2}}{20 - 4 x + x ^{2}} d x

What is the solution for y^{''}+1.5y^'-y=12x^2+6x^3-x^4 ?
The solution for y^{''}+1.5y^'-y=12x^2+6x^3-x^4 is y=c_{1}e^{x/2}+c_{2}e^{-2x}+x^4