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Popular >

16y^{''}-40y^'+25y=0,y(4)=2,y^'(4)=6

Solution

16 y^{^{''}} - 40 y^{^{'}} + 25 y = 0, y (4) = 2, y^{^{'}} (4) = 6

Solution

y = - 12 e^{\frac{5 t - 20}{4}} + \frac{7 e ^{\frac{5 t - 20}{4}} t}{2}

Solution steps

16 y^{''} (t) - 40 y^{'} (t) + 25 y = 0

Solve linear ODE:

y = - 12 e^{\frac{5 t - 20}{4}} + \frac{7 e ^{\frac{5 t - 20}{4}} t}{2}

y = - 12 e^{\frac{5 t - 20}{4}} + \frac{7 e ^{\frac{5 t - 20}{4}} t}{2}

Graph

Popular Examples

y^{''}+2y^'+10y=0,y(pi/2)=0

y^{^{''}} + 2 y^{^{'}} + 10 y = 0, y (\frac{π}{2}) = 0

y^{''}-2y^'+y=0,y(0)=4,y^'(0)=-4

y^{^{''}} - 2 y^{^{'}} + y = 0, y (0) = 4, y^{^{'}} (0) = - 4

y^{''}+4y^'+4y=0,y(-1)=3,y^'(-2)=5

y^{^{''}} + 4 y^{^{'}} + 4 y = 0, y (- 1) = 3, y^{^{'}} (- 2) = 5

y^{''}+8y^'+25y=0,y(0)=4,y^'(0)=-15

y^{^{''}} + 8 y^{^{'}} + 25 y = 0, y (0) = 4, y^{^{'}} (0) = - 15

6(d^2y)/(dx^2)+2(dy)/(dx)-y=0

6 \frac{d ^{2} y}{d x ^{2}} + 2 \frac{d y}{d x} - y = 0

Frequently Asked Questions (FAQ)

What is the solution for 16y^{''}-40y^'+25y=0,y(4)=2,y^'(4)=6 ?
The solution for 16y^{''}-40y^'+25y=0,y(4)=2,y^'(4)=6 is y=-12e^{(5t-20)/4}+(7e^{(5t-20)/4}t)/(2)