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Popular >

y^'-2y=-1,y(0)=3

Solution

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

Solution

y = \frac{1}{2} + \frac{5}{2} e^{2 t}

Solution steps

y^{'} (t) - 2 y = - 1

Solve linear ODE:

y = \frac{1}{2} + \frac{5}{2} e^{2 t}

y = \frac{1}{2} + \frac{5}{2} e^{2 t}

Popular Examples

(y-x)e^x+(1+e^x)y^'=0

(y - x) e^{x} + (1 + e^{x}) y^{^{'}} = 0

derivative of 10+(50t)/(2t^2+9)derivative of

10 + \frac{50 t}{2 t ^{2} + 9}

limit as x approaches-2 of 1/2

x \to - 2 lim (\frac{1}{2})

inverse oflaplace (-3e^{6-3s})/((s-2)(s-3)(s+2))inverse laplace

\frac{- 3 e ^{6 - 3 s}}{( s - 2 ) ( s - 3 ) ( s + 2 )}

limit as h approaches 0 of (x+h)^3-x^3

h \to 0 lim ((x + h)^{3} - x^{3})

Frequently Asked Questions (FAQ)

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