What is the solution for (y^{''}-ay^')=by ?

The solution for (y^{''}-ay^')=by is y=c_{1}e^{(t(a+sqrt(a^2+4b)))/2}+c_{2}e^{(t(a-sqrt(a^2+4b)))/2}

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

(y^{''}-ay^')=by

(y^{^{''}} - a y^{^{'}}) = b y

y = c_{1} e^{\frac{t ( a + a ^{2} + 4 b )}{2}} + c_{2} e^{\frac{t ( a - a ^{2} + 4 b )}{2}}

Solution steps

(y^{''} (t) - a y^{'} (t)) = b y (t)

Solve linear ODE:

y = c_{1} e^{\frac{t ( a + a ^{2} + 4 b )}{2}} + c_{2} e^{\frac{t ( a - a ^{2} + 4 b )}{2}}

y = c_{1} e^{\frac{t ( a + a ^{2} + 4 b )}{2}} + c_{2} e^{\frac{t ( a - a ^{2} + 4 b )}{2}}

y^{^{''}} + 12 y^{^{'}} + 4 y = 0, y (0) = - 3, y^{^{'}} (0) = 0

y^{^{''}} + 16.2 y^{^{'}} = 0

(D^{2} + 4 D + 4) y = 0

y^{^{''}} - 7 y^{^{'}} - 98 y = 0

\frac{d ^{2} y}{d x ^{2}} - 3 \frac{d y}{d x} + 4 y = 0

What is the solution for (y^{''}-ay^')=by ?
The solution for (y^{''}-ay^')=by is y=c_{1}e^{(t(a+sqrt(a^2+4b)))/2}+c_{2}e^{(t(a-sqrt(a^2+4b)))/2}