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Popular >

y^{''}+16y=t^2+1,y(0)=0,y^'(0)=1

Solution

y^{^{''}} + 16 y = t^{2} + 1, y (0) = 0, y^{^{'}} (0) = 1

Solution

y = - \frac{7}{128} cos (4 t) + \frac{1}{4} sin (4 t) + \frac{t ^{2}}{16} + \frac{7}{128}

Solution steps

y^{''} (t) + 16 y = t^{2} + 1

Solve linear ODE:

y = - \frac{7}{128} cos (4 t) + \frac{1}{4} sin (4 t) + \frac{t ^{2}}{16} + \frac{7}{128}

y = - \frac{7}{128} cos (4 t) + \frac{1}{4} sin (4 t) + \frac{t ^{2}}{16} + \frac{7}{128}

Graph

Popular Examples

y^{'''}-12y^{''}+48y^'-64y=9e^{4x}

y^{^{'''}} - 12 y^{^{''}} + 48 y^{^{'}} - 64 y = 9 e^{4 x}

y^{'''}+121y^'=tan(11x)

y^{^{'''}} + 121 y^{^{'}} = tan (11 x)

y^{''}-2y^'-3y=6xe^{2x}+4x-5

y^{^{''}} - 2 y^{^{'}} - 3 y = 6 x e^{2 x} + 4 x - 5

y^{''}-y^'-2y=sin(3t)

y^{^{''}} - y^{^{'}} - 2 y = sin (3 t)

y^{''}+6y=4t^2-3,y(0)=0,y^'(0)=-7

y^{^{''}} + 6 y = 4 t^{2} - 3, y (0) = 0, y^{^{'}} (0) = - 7

Frequently Asked Questions (FAQ)

What is the solution for y^{''}+16y=t^2+1,y(0)=0,y^'(0)=1 ?
The solution for y^{''}+16y=t^2+1,y(0)=0,y^'(0)=1 is y=-7/128 cos(4t)+1/4 sin(4t)+(t^2)/(16)+7/128