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Popular >

y^{''}-5y^'+4y=e^{2t},y(0)=1,y^'(0)=-1

Solution

y^{^{''}} - 5 y^{^{'}} + 4 y = e^{2 t}, y (0) = 1, y^{^{'}} (0) = - 1

Solution

y = (- 2 + \frac{3}{2}) e^{4 t} + 2 e^{t} - \frac{1}{2} e^{2 t}

Solution steps

y^{''} (t) - 5 y^{'} (t) + 4 y = e^{2 t}

Solve linear ODE:

y = (- 2 + \frac{3}{2}) e^{4 t} + 2 e^{t} - \frac{1}{2} e^{2 t}

y = (- 2 + \frac{3}{2}) e^{4 t} + 2 e^{t} - \frac{1}{2} e^{2 t}

Popular Examples

y^{''}+4y=4sin(2t),y(pi)=0,y^'(pi)=-1

y^{^{''}} + 4 y = 4 sin (2 t), y (π) = 0, y^{^{'}} (π) = - 1

2y^{''}+8y^'+80y=40,y(0)=0,y^'(0)=0

2 y^{^{''}} + 8 y^{^{'}} + 80 y = 40, y (0) = 0, y^{^{'}} (0) = 0

y^{''}+16y=12cos(4x)+18sin(4x)

y^{^{''}} + 16 y = 12 cos (4 x) + 18 sin (4 x)

linear-1/4 y^{''}+y^'+y=x^2-2x linear

- \frac{1}{4} y^{^{''}} + y^{^{'}} + y = x^{2} - 2 x

y^{''}-10y^'+25y=(e^{5x})/(4+x^2)

y^{^{''}} - 10 y^{^{'}} + 25 y = \frac{e ^{5 x}}{4 + x ^{2}}

Frequently Asked Questions (FAQ)

What is the solution for y^{''}-5y^'+4y=e^{2t},y(0)=1,y^'(0)=-1 ?
The solution for y^{''}-5y^'+4y=e^{2t},y(0)=1,y^'(0)=-1 is y=(-2+3/2)e^{4t}+2e^t-1/2 e^{2t}