What is the solution for (d^2y)/(dx^2)=0.15y,y(0)=240,y(10)=150 ?

The solution for (d^2y)/(dx^2)=0.15y,y(0)=240,y(10)=150 is

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(d^2y)/(dx^2)=0.15y,y(0)=240,y(10)=150

\frac{d ^{2} y}{d x ^{2}} = 0.15 y, y (0) = 240, y (10) = 150

y = (240 - \frac{- 150 e ^{15} + 240 e ^{215}}{e ^{215} - 1}) e^{\frac{3 x}{2 5}} + \frac{- 150 e ^{15} + 240 e ^{215}}{e ^{215} - 1} e^{- \frac{3 x}{2 5}}

Solution steps

\frac{d ^{2} y}{d x ^{2}} = 0.15 y

Solve linear ODE:

y = (240 - \frac{- 150 e ^{15} + 240 e ^{215}}{e ^{215} - 1}) e^{\frac{3 x}{2 5}} + \frac{- 150 e ^{15} + 240 e ^{215}}{e ^{215} - 1} e^{- \frac{3 x}{2 5}}

y = (240 - \frac{- 150 e ^{15} + 240 e ^{215}}{e ^{215} - 1}) e^{\frac{3 x}{2 5}} + \frac{- 150 e ^{15} + 240 e ^{215}}{e ^{215} - 1} e^{- \frac{3 x}{2 5}}

y^{^{''}} - 4 y^{^{'}} + 13 y = 0, y (0) = - 2, y^{^{'}} (0) = - 16

y = 10 y^{^{''}}

y^{^{''}} - 2 y^{^{'}} - 6 y = 0

y^{^{''}} + 3 y^{^{'}} - 18 y = 0, y (0) = - 1, y^{^{'}} (0) = 24

y^{^{''}} + 4 y^{^{'}} + 5 y = 0, y (0) = 1

What is the solution for (d^2y)/(dx^2)=0.15y,y(0)=240,y(10)=150 ?
The solution for (d^2y)/(dx^2)=0.15y,y(0)=240,y(10)=150 is