What is the solution for y^{''}-7y^'+10y=0,y^'(0)=-1/3 ?

The solution for y^{''}-7y^'+10y=0,y^'(0)=-1/3 is y=(e^{5t}(-1-6c_{2}))/(15)+c_{2}e^{2t}

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^{''}-7y^'+10y=0,y^'(0)=-1/3

y^{^{''}} - 7 y^{^{'}} + 10 y = 0, y^{^{'}} (0) = - \frac{1}{3}

y = \frac{e ^{5 t} ( - 1 - 6 c _{2} )}{15} + c_{2} e^{2 t}

Solution steps

y^{''} (t) - 7 y^{'} (t) + 10 y = 0

Solve linear ODE:

y = \frac{e ^{5 t} ( - 1 - 6 c _{2} )}{15} + c_{2} e^{2 t}

y = \frac{e ^{5 t} ( - 1 - 6 c _{2} )}{15} + c_{2} e^{2 t}

(D^{2} - 6 D + 25) y = 0

4 y^{^{''}} + 4 y^{^{'}} + y = 0, y (0) = - 1, y^{^{'}} (0) = 2

\frac{d ^{2} y}{d x ^{2}} + 36 y = 0

x^{^{''}} - 2 x^{^{'}} + 5 x = 0

y^{^{''}} + 49 y = 0, y (0) = 5, y^{^{'}} (0) = - 7

What is the solution for y^{''}-7y^'+10y=0,y^'(0)=-1/3 ?
The solution for y^{''}-7y^'+10y=0,y^'(0)=-1/3 is y=(e^{5t}(-1-6c_{2}))/(15)+c_{2}e^{2t}