What is the solution for -2y^{''}+9y=0,y(-2)=2,y^'(2)=1 ?

The solution for -2y^{''}+9y=0,y(-2)=2,y^'(2)=1 is

Popular >

-2y^{''}+9y=0,y(-2)=2,y^'(2)=1

- 2 y^{^{''}} + 9 y = 0, y (- 2) = 2, y^{^{'}} (2) = 1

y = 2 e^{32} - \frac{e ^{62} ( 6 e ^{92} - 2 e ^{32} )}{3 ( e ^{122} + 1 )} e^{\frac{3 t}{2}} + \frac{6 e ^{92} - 2 e ^{32}}{3 ( e ^{122} + 1 )} e^{- \frac{3 t}{2}}

Solution steps

- 2 y^{''} (t) + 9 y = 0

Solve linear ODE:

y = 2 e^{32} - \frac{e ^{62} ( 6 e ^{92} - 2 e ^{32} )}{3 ( e ^{122} + 1 )} e^{\frac{3 t}{2}} + \frac{6 e ^{92} - 2 e ^{32}}{3 ( e ^{122} + 1 )} e^{- \frac{3 t}{2}}

y = 2 e^{32} - \frac{e ^{62} ( 6 e ^{92} - 2 e ^{32} )}{3 ( e ^{122} + 1 )} e^{\frac{3 t}{2}} + \frac{6 e ^{92} - 2 e ^{32}}{3 ( e ^{122} + 1 )} e^{- \frac{3 t}{2}}

\frac{d}{d x} (in (x + 1 + x^{2}))

f (x) = 2 x^{3} - x^{2} + 1, (1, 2)

\frac{( 4 x - 2 ) ^{4}}{( - 5 x + 1 ) ^{5}}

f (x) = (\frac{1}{x - 3})^{2}

x^{3} + 2 x^{2} + 1

What is the solution for -2y^{''}+9y=0,y(-2)=2,y^'(2)=1 ?
The solution for -2y^{''}+9y=0,y(-2)=2,y^'(2)=1 is