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Popular >

y^'+ty=ty^2

Solution

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

Solution

y = \frac{1}{1 - e ^{\frac{t ^{2}}{2} + c_{1}}}

Solution steps

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

Solve separable ODE:

y = \frac{1}{1 - e ^{\frac{t ^{2}}{2} + c_{1}}}

y = \frac{1}{1 - e ^{\frac{t ^{2}}{2} + c_{1}}}

Popular Examples

y^{''}-4y=(4x)/(e^{2x)}

y^{^{''}} - 4 y = \frac{4 x}{e ^{2 x}}

sum from n=0 to infinity of 1/(n+e^n)

n = 0 \sum \infty \frac{1}{n + e ^{n}}

derivative of (24/((x+2)^3))

\frac{d}{d x} (\frac{24}{( x + 2 ) ^{3}})

x^2+3xy+x(dy)/(dx)=0

x^{2} + 3 x y + x \frac{d y}{d x} = 0

area y=16-x^2,y=0,x=-2,x=3 area

y = 16 - x^{2}, y = 0, x = - 2, x = 3

Frequently Asked Questions (FAQ)

What is the solution for y^'+ty=ty^2 ?
The solution for y^'+ty=ty^2 is y= 1/(1-e^{\frac{t^2){2}+c_{1}}}