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Popular >

ty^'+2y=4t^2,y(1)=2

Solution

t y^{^{'}} + 2 y = 4 t^{2}, y (1) = 2

Solution

y = t^{2} + \frac{1}{t ^{2}}

Solution steps

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

Solve linear ODE:

y = t^{2} + \frac{1}{t ^{2}}

y = t^{2} + \frac{1}{t ^{2}}

Graph

Popular Examples

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\int e^{z} cos (x) d x

tangent of f(x)=-1-4x^2,(3,-37)tangent of

f (x) = - 1 - 4 x^{2}, (3, - 37)

2 y^{^{''}} + 72 y = 0

d/(dy)(ln(cos(y)))

\frac{d}{d y} (ln (cos (y)))

Frequently Asked Questions (FAQ)

What is the solution for ty^'+2y=4t^2,y(1)=2 ?
The solution for ty^'+2y=4t^2,y(1)=2 is y=t^2+1/(t^2)