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Popular >

xy^2+yy^'=x^2yy^'

Solution

x y^{2} + y y^{^{'}} = x^{2} y y^{^{'}}

Solution

y = c_{1} x^{2} - 1

Solution steps

x y^{2} + y y^{'} (x) = x^{2} y y^{'} (x)

Solve separable ODE:

y = c_{1} x^{2} - 1

y = c_{1} x^{2} - 1

Popular Examples

(x^2}{y^2-1}(dy)/(dx)=\frac{3x^3)/y

\frac{x ^{2}}{y ^{2} - 1} \frac{d y}{d x} = \frac{3 x ^{3}}{y}

(dQ)/(dt)+0.05Q=0

\frac{d Q}{d t} + 0.05 Q = 0

y^'=(2+3x)sqrt(y),y(0)=3

y^{^{'}} = (2 + 3 x) y, y (0) = 3

y^'(s)=((1-2s)/(s^2))y(s)

y^{^{'}} (s) = (\frac{1 - 2 s}{s ^{2}}) y (s)

y^'+(cos(x))y=5cos(x)

y^{^{'}} + (cos (x)) y = 5 cos (x)

Frequently Asked Questions (FAQ)

What is the solution for xy^2+yy^'=x^2yy^' ?
The solution for xy^2+yy^'=x^2yy^' is y=c_{1}sqrt(x^2-1)