What is the solution for y^{''}+4y^'-5y=0,y^'(ln(2))=3,y(0)=33 ?

The solution for y^{''}+4y^'-5y=0,y^'(ln(2))=3,y(0)=33 is y=(87e^t)/(23)+672/23 e^{-5t}

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y^{''}+4y^'-5y=0,y^'(ln(2))=3,y(0)=33

y^{^{''}} + 4 y^{^{'}} - 5 y = 0, y^{^{'}} (ln (2)) = 3, y (0) = 33

y = \frac{87 e ^{t}}{23} + \frac{672}{23} e^{- 5 t}

Solution steps

y^{''} (t) + 4 y^{'} (t) - 5 y = 0

Solve linear ODE:

y = \frac{87 e ^{t}}{23} + \frac{672}{23} e^{- 5 t}

y = \frac{87 e ^{t}}{23} + \frac{672}{23} e^{- 5 t}

\frac{d ^{2} x}{d t ^{2}} - 49 x = 0, x = 2

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

- 3 y^{^{''}} + 27 y = 0, y^{^{'}} (- 2) = - 2, y (2) = - 2

y^{^{''}} + 5 y^{^{'}} = 0, y (0) = - 11, y^{^{'}} (0) = 35

12 y^{^{''}} - 11 y^{^{'}} + 2 y = 0, y (0) = 7, y^{^{'}} (0) = - 3

What is the solution for y^{''}+4y^'-5y=0,y^'(ln(2))=3,y(0)=33 ?
The solution for y^{''}+4y^'-5y=0,y^'(ln(2))=3,y(0)=33 is y=(87e^t)/(23)+672/23 e^{-5t}