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Popular >

y^{''}+4y=t^2+10e^t,y(0)=0,y^'(0)=3

Solution

y^{^{''}} + 4 y = t^{2} + 10 e^{t}, y (0) = 0, y^{^{'}} (0) = 3

Solution

y = - \frac{15}{8} cos (2 t) + \frac{1}{2} sin (2 t) + \frac{t ^{2}}{4} - \frac{1}{8} + 2 e^{t}

Solution steps

y^{''} (t) + 4 y = t^{2} + 10 e^{t}

Solve linear ODE:

y = - \frac{15}{8} cos (2 t) + \frac{1}{2} sin (2 t) + \frac{t ^{2}}{4} - \frac{1}{8} + 2 e^{t}

y = - \frac{15}{8} cos (2 t) + \frac{1}{2} sin (2 t) + \frac{t ^{2}}{4} - \frac{1}{8} + 2 e^{t}

Popular Examples

y^{''}+y=4x*sin(x)

y^{^{''}} + y = 4 x \cdot sin (x)

y^{''}-10y^'+25y=(1.5e^{5t})/(t^2+1)

y^{^{''}} - 10 y^{^{'}} + 25 y = \frac{1.5 e ^{5 t}}{t ^{2} + 1}

(d^2y)/(dx^2)-4(dy)/(dx)+12y=3cos(x)

\frac{d ^{2} y}{d x ^{2}} - 4 \frac{d y}{d x} + 12 y = 3 cos (x)

y^{'''}-6y^{''}+12y^'-8y=2e^{2x}

y^{^{'''}} - 6 y^{^{''}} + 12 y^{^{'}} - 8 y = 2 e^{2 x}

y^{''}-3y^'=1+e^x+cos(x)+sin(x)

y^{^{''}} - 3 y^{^{'}} = 1 + e^{x} + cos (x) + sin (x)

Frequently Asked Questions (FAQ)

What is the solution for y^{''}+4y=t^2+10e^t,y(0)=0,y^'(0)=3 ?
The solution for y^{''}+4y=t^2+10e^t,y(0)=0,y^'(0)=3 is y=-15/8 cos(2t)+1/2 sin(2t)+(t^2)/4-1/8+2e^t