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Popular >

y^'+4y=1,y(0)=1

Solution

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

Solution

y = \frac{1}{4} + \frac{3}{4 e ^{4 t}}

Solution steps

y^{'} (t) + 4 y = 1

Solve linear ODE:

y = \frac{1}{4} + \frac{3}{4 e ^{4 t}}

y = \frac{1}{4} + \frac{3}{4 e ^{4 t}}

Popular Examples

tangent of f(x)=sqrt(x^2+16),\at x=-3 tangent of

f (x) = x^{2} + 16, at x = - 3

expand (x+5)cos(x)expand

(x + 5) cos (x)

y^{''}+4y^'+9y=0

y^{^{''}} + 4 y^{^{'}} + 9 y = 0

integral of (x^2)/(sqrt(196-x^2))

\int \frac{x ^{2}}{196 - x ^{2}} d x

y^{''}+2y^'+37y=0,y(0)=-3,y^'(0)=21

y^{^{''}} + 2 y^{^{'}} + 37 y = 0, y (0) = - 3, y^{^{'}} (0) = 21

Frequently Asked Questions (FAQ)

What is the solution for y^'+4y=1,y(0)=1 ?
The solution for y^'+4y=1,y(0)=1 is y= 1/4+3/(4e^{4t)}