en

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

4y^{''}-y^'+6=0

Solution

4 y^{^{''}} - y^{^{'}} + 6 = 0

Solution

y = c_{1} e^{\frac{t}{4}} + c_{2} + 6 t

Solution steps

4 y^{''} (t) - y^{'} (t) + 6 = 0

Solve linear ODE:

y = c_{1} e^{\frac{t}{4}} + c_{2} + 6 t

y = c_{1} e^{\frac{t}{4}} + c_{2} + 6 t

Graph

Popular Examples

y^{^{''}} + 4 y = 2 e^{t}

(d^2y)/(dx^2)+2(dy)/(dx)+2*y=2

\frac{d ^{2} y}{d x ^{2}} + 2 \frac{d y}{d x} + 2 \cdot y = 2

y^{''}+6y^'+9y=((16e^{-3x}))/(x^2+1)

y^{^{''}} + 6 y^{^{'}} + 9 y = \frac{( 16 e ^{- 3 x} )}{x ^{2} + 1}

y^{''}-2y^'-24y=72e^{2t}

y^{^{''}} - 2 y^{^{'}} - 24 y = 72 e^{2 t}

y^{''}+49y=t^2cos(7t)+te^{(-7t)}

y^{^{''}} + 49 y = t^{2} cos (7 t) + t e^{(- 7 t)}

Frequently Asked Questions (FAQ)

What is the solution for 4y^{''}-y^'+6=0 ?
The solution for 4y^{''}-y^'+6=0 is y=c_{1}e^{t/4}+c_{2}+6t