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Popular >

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

Solution

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

Solution

y = c_{1} e^{2 t} + c_{2} e^{t} + c_{3} t e^{t} - \frac{t ^{2}}{2} - \frac{5 t}{2} - \frac{17}{4}

Solution steps

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

Solve linear ODE:

y = c_{1} e^{2 t} + c_{2} e^{t} + c_{3} t e^{t} - \frac{t ^{2}}{2} - \frac{5 t}{2} - \frac{17}{4}

y = c_{1} e^{2 t} + c_{2} e^{t} + c_{3} t e^{t} - \frac{t ^{2}}{2} - \frac{5 t}{2} - \frac{17}{4}

Graph

Popular Examples

y^{''}+4y=3cos(t),y(0)=0,y^'(0)=0

y^{^{''}} + 4 y = 3 cos (t), y (0) = 0, y^{^{'}} (0) = 0

y^{''}-3y^'-10=0,y(0)=1,y^'(0)=10

y^{^{''}} - 3 y^{^{'}} - 10 = 0, y (0) = 1, y^{^{'}} (0) = 10

2y^{''}+5y^'-y=9

2 y^{^{''}} + 5 y^{^{'}} - y = 9

y^{''}+7y^'+12y=340cos(t)

y^{^{''}} + 7 y^{^{'}} + 12 y = 340 cos (t)

y^{''}+y=2,+ty(0)=1,y(0)=1

y^{^{''}} + y = 2, + t y (0) = 1, y (0) = 1

Frequently Asked Questions (FAQ)

What is the solution for y^{'''}-4y^{''}+5y^'-2y=t^2 ?
The solution for y^{'''}-4y^{''}+5y^'-2y=t^2 is y=c_{1}e^{2t}+c_{2}e^t+c_{3}te^t-(t^2)/2-(5t)/2-17/4