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Solve this initial value problem:

$$5y^{(3)}+ 2y'' = 0 , y(0) = 4 , y'(0) = 2, y''(0) = 8$$

The only method we have learned is to assign variables to each degree, and then make a matrix to solve the equation. So, for this problem, would I do something like this?

$$a = y, b = y', c = y'' \\ a' = b\\ b' = c\\ c' = -2c/5$$

And then plug into the matrix? Note: the $y^{(3)}$ is a third derivative; I tried formatting and it didn't work

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    **Hint:** What if we write $5 m^3 + 2 m^2 = 0 \implies m_{1,2,3} = 0, 0, -\dfrac{2}{5}$? We can now write $y_c(x) = c_1 + c_2 x + c_3 e^{-2/5 x}$.Now, use your ICs to find the constants. – 2017-02-19

2 Answers 2

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As long as the equation has constant coefficients the solutions have the form $y = e^{rx}$. If you plug this into the differential equation you get $5r^3 + 2r^2=0$, which has a root of $-2/5$ and a repeated root of $0$. The general solution is a linear combination of $e^{0x}=1$, $xe^{0x}=x$, and $e^{-2x/5}$.

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Or you can do it like this $$5y^{(3)}+2y''=0$$ Let $y''=g(x)$ then $$5g'+2g=0$$ So, $$g(x)=ce^{-2x/5}$$ and $g(0)=8$, Hence $$y''(x)=8e^{-2x/5}$$ $$y'(x)=c-20e^{-2x/5}$$ $$y'(0)=2$$ Hence $$y'(x)=22-20e^{-2x/5}$$ $$y(x)=c+22x+50e^{-2x/5}$$ $$y(0)=4$$ So $$y(x)=50e^{-2x/5}+22x-46$$