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Popular >

y^{''}-10y^'+26y=4,y(0)=3,y^'(0)=15

Solution

y^{^{''}} - 10 y^{^{'}} + 26 y = 4, y (0) = 3, y^{^{'}} (0) = 15

Solution

y = e^{5 t} (\frac{37}{13} cos (t) + \frac{10}{13} sin (t)) + \frac{2}{13}

Solution steps

y^{''} (t) - 10 y^{'} (t) + 26 y = 4

Solve linear ODE:

y = e^{5 t} (\frac{37}{13} cos (t) + \frac{10}{13} sin (t)) + \frac{2}{13}

y = e^{5 t} (\frac{37}{13} cos (t) + \frac{10}{13} sin (t)) + \frac{2}{13}

Graph

Popular Examples

(d^2y)/(dx^2)-2(dy)/(dx)+5y=e^xcos(2x)

\frac{d ^{2} y}{d x ^{2}} - 2 \frac{d y}{d x} + 5 y = e^{x} cos (2 x)

y^{'''}-y^{''}-y^'+y=2e^{-2x}+3

y^{^{'''}} - y^{^{''}} - y^{^{'}} + y = 2 e^{- 2 x} + 3

x^{''}-4x^'+4x=e^{2t},x(0)=2,x^'(0)=1

x^{^{''}} - 4 x^{^{'}} + 4 x = e^{2 t}, x (0) = 2, x^{^{'}} (0) = 1

y^{'''}+256y^'=tan(16x)

y^{^{'''}} + 256 y^{^{'}} = tan (16 x)

y^{''}-4y=3x^2e^{3x}

y^{^{''}} - 4 y = 3 x^{2} e^{3 x}

Frequently Asked Questions (FAQ)

What is the solution for y^{''}-10y^'+26y=4,y(0)=3,y^'(0)=15 ?
The solution for y^{''}-10y^'+26y=4,y(0)=3,y^'(0)=15 is y=e^{5t}(37/13 cos(t)+10/13 sin(t))+2/13