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Popular >

t^2y^'+y^'=ty

Solution

t^{2} \cdot y^{^{'}} + y^{^{'}} = t \cdot y

Solution

y = c_{1} t^{2} + 1

Solution steps

t^{2} \cdot y^{'} (t) + y^{'} (t) = t \cdot y

Solve separable ODE:

y = c_{1} t^{2} + 1

y = c_{1} t^{2} + 1

Popular Examples

derivative of (2x-3/(2x+3))

\frac{d}{d x} (\frac{2 x - 3}{2 x + 3})

y^{'''}+y^{''}-y^'-y=4e^x

y^{^{'''}} + y^{^{''}} - y^{^{'}} - y = 4 e^{x}

derivative of sin^2(x-3)

\frac{d}{d x} (sin^{2} (x) - 3)

(\partial)/(\partial x)(e^{5x}cos(7y))

\frac{\partial}{\partial x} (e^{5 x} cos (7 y))

derivative of (2x+3/(5x+4))

\frac{d}{d x} (\frac{2 x + 3}{5 x + 4})

Frequently Asked Questions (FAQ)

What is the solution for t^2*y^'+y^'=t*y ?
The solution for t^2*y^'+y^'=t*y is y=c_{1}sqrt(t^2+1)