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Popular >

2yt(dy)+y^2(dt)=0

Solution

2 y t (d y) + y^{2} (d t) = 0

Solution

y = \frac{c _{1}}{t}

Solution steps

2 y t (d y) + y^{2} (d t) = 0

Solve separable ODE:

y = \frac{c _{1}}{t}

y = \frac{c _{1}}{t}

Popular Examples

y^'+y=y(te^{t^2}+1),y(0)=1

y^{^{'}} + y = y (t e^{t^{2}} + 1), y (0) = 1

y^'=(y^2)/(x^3),y(2)=4

y^{^{'}} = \frac{y ^{2}}{x ^{3}}, y (2) = 4

xyy^'=(x+1)(y+1)

x y y^{^{'}} = (x + 1) (y + 1)

(3y^2-4x^{-1}y)dx+(3xy-4)dy=0

(3 y^{2} - 4 x^{- 1} y) d x + (3 x y - 4) d y = 0

x^{^{'}} = \frac{x ^{2}}{2}

Frequently Asked Questions (FAQ)

What is the solution for 2yt(dy)+y^2(dt)=0 ?
The solution for 2yt(dy)+y^2(dt)=0 is y=(c_{1})/(sqrt(t))