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Popular >

t^2x^'+x=0

Solution

t^{2} x^{^{'}} + x = 0

Solution

x = e^{\frac{1}{t} + c_{1}}

Solution steps

t^{2} x^{'} (t) + x = 0

Solve separable ODE:

x = e^{\frac{1}{t} + c_{1}}

x = e^{\frac{1}{t} + c_{1}}

Popular Examples

(dy)/(dx)=(e^{5x})/(cos(2y))

\frac{d y}{d x} = \frac{e ^{5 x}}{cos ( 2 y )}

y^{^{'}} - 2 x^{2} y - y = 0

y^'=(e^x)/(e^y)

y^{^{'}} = \frac{e ^{x}}{e ^{y}}

yy^'=[sin(x)-cos(x)]sqrt(1-y^2)

y y^{^{'}} = [sin (x) - cos (x)] 1 - y^{2}

(dy)/(dx)=13xsec(y)

\frac{d y}{d x} = 13 x sec (y)

Frequently Asked Questions (FAQ)

What is the solution for t^2x^'+x=0 ?
The solution for t^2x^'+x=0 is x=e^{1/t+c_{1}}