en

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^'=2y+3,y(0)=5

Solution

y^{^{'}} = 2 y + 3, y (0) = 5

Solution

y = \frac{13 e ^{2 t}}{2} - \frac{3}{2}

Solution steps

y^{'} (t) = 2 y + 3

Solve separable ODE:

y = \frac{13 e ^{2 t}}{2} - \frac{3}{2}

y = \frac{13 e ^{2 t}}{2} - \frac{3}{2}

Graph

Popular Examples

((dy}{dx})=\frac{2y)/x

(\frac{d y}{d x}) = \frac{2 y}{x}

(dy}{dt}=\frac{sin^2(t)+t)/y

\frac{d y}{d t} = \frac{sin ^{2} ( t ) + t}{y}

2 (x - 1) y^{^{'}} - 3 y = 0

y^'(x)=(x^2)/(1-y^2),y(2)=3

y^{^{'}} (x) = \frac{x ^{2}}{1 - y ^{2}}, y (2) = 3

(xy^2+4y^2)dy-5xdx=0

(x y^{2} + 4 y^{2}) d y - 5 x d x = 0

Frequently Asked Questions (FAQ)

What is the solution for y^'=2y+3,y(0)=5 ?
The solution for y^'=2y+3,y(0)=5 is y=(13e^{2t})/2-3/2