What is the solution for 36y^{''}-y=0,y(-6)=1,y^'(-6)=-1 ?

The solution for 36y^{''}-y=0,y(-6)=1,y^'(-6)=-1 is y=(e-(7e)/2)e^{t/6}+(7e^{(-t-6)/6})/(2)

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36y^{''}-y=0,y(-6)=1,y^'(-6)=-1

36 y^{^{''}} - y = 0, y (- 6) = 1, y^{^{'}} (- 6) = - 1

y = (e - \frac{7 e}{2}) e^{\frac{t}{6}} + \frac{7 e ^{\frac{- t - 6}{6}}}{2}

Solution steps

36 y^{''} (t) - y = 0

Solve linear ODE:

y = (e - \frac{7 e}{2}) e^{\frac{t}{6}} + \frac{7 e ^{\frac{- t - 6}{6}}}{2}

y = (e - \frac{7 e}{2}) e^{\frac{t}{6}} + \frac{7 e ^{\frac{- t - 6}{6}}}{2}

x \to - 5 lim (- 6 x)

(2 x y - 3 x^{2}) d x + (x^{2} + y) d y = 0

\frac{\partial}{\partial y} (cos (x y))

\int_{0}^{3} 1 + x^{4} d x

x \to 5 lim (2 x^{2})

What is the solution for 36y^{''}-y=0,y(-6)=1,y^'(-6)=-1 ?
The solution for 36y^{''}-y=0,y(-6)=1,y^'(-6)=-1 is y=(e-(7e)/2)e^{t/6}+(7e^{(-t-6)/6})/(2)