What is the solution for 7y^{''}=5y-3/2 x^2,y(0)=0,y^'(2)=0 ?

The solution for 7y^{''}=5y-3/2 x^2,y(0)=0,y^'(2)=0 is

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7y^{''}=5y-3/2 x^2,y(0)=0,y^'(2)=0

7 y^{^{''}} = 5 y - \frac{3}{2} x^{2}, y (0) = 0, y^{^{'}} (2) = 0

y = \frac{- 6 35 e ^{\frac{2 5}{7}} + 21 e ^{\frac{4 5}{7}}}{25 ( e ^{\frac{4 5}{7}} + 1 )} - \frac{21}{25} e^{\frac{5}{7} x} - \frac{- 6 35 e ^{\frac{2 5}{7}} + 21 e ^{\frac{4 5}{7}}}{25 ( e ^{\frac{4 5}{7}} + 1 )} e^{- \frac{5}{7} x} + \frac{3 x ^{2}}{10} + \frac{21}{25}

Solution steps

7 y^{''} (x) = 5 y - \frac{3}{2} x^{2}

Solve linear ODE:

y = \frac{- 6 35 e ^{\frac{2 5}{7}} + 21 e ^{\frac{4 5}{7}}}{25 ( e ^{\frac{4 5}{7}} + 1 )} - \frac{21}{25} e^{\frac{5}{7} x} - \frac{- 6 35 e ^{\frac{2 5}{7}} + 21 e ^{\frac{4 5}{7}}}{25 ( e ^{\frac{4 5}{7}} + 1 )} e^{- \frac{5}{7} x} + \frac{3 x ^{2}}{10} + \frac{21}{25}

y = \frac{- 6 35 e ^{\frac{2 5}{7}} + 21 e ^{\frac{4 5}{7}}}{25 ( e ^{\frac{4 5}{7}} + 1 )} - \frac{21}{25} e^{\frac{5}{7} x} - \frac{- 6 35 e ^{\frac{2 5}{7}} + 21 e ^{\frac{4 5}{7}}}{25 ( e ^{\frac{4 5}{7}} + 1 )} e^{- \frac{5}{7} x} + \frac{3 x ^{2}}{10} + \frac{21}{25}

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

\frac{d ^{2} y}{d x ^{2}} - 4 \frac{d y}{d x} + 3 y = e^{4 x}

y^{^{''}} + 8 y^{^{'}} - 9 y = 64 x + 1

\frac{d ^{2} y}{d x ^{2}} + y = 3 e^{- 5 x}

y^{^{''}} - 2 y^{^{'}} + 5 y = cos (2 x)

What is the solution for 7y^{''}=5y-3/2 x^2,y(0)=0,y^'(2)=0 ?
The solution for 7y^{''}=5y-3/2 x^2,y(0)=0,y^'(2)=0 is