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Popular >

y^'=y^2(9x+5)

Solution

y^{^{'}} = y^{2} (9 x + 5)

Solution

y = - \frac{2}{9 x ^{2} + 10 x + c _{1}}

Solution steps

y^{'} (x) = y^{2} (9 x + 5)

Solve separable ODE:

y = - \frac{2}{9 x ^{2} + 10 x + c _{1}}

y = - \frac{2}{9 x ^{2} + 10 x + c _{1}}

Popular Examples

y^'sec(x)=5y^2,y(pi/2)= 1/2

y^{^{'}} sec (x) = 5 y^{2}, y (\frac{π}{2}) = \frac{1}{2}

(dQ)/(dt)+Q/(20)=0

\frac{d Q}{d t} + \frac{Q}{20} = 0

(dy)/(dx)xy+2x+3x^2=4

\frac{d y}{d x} x y + 2 x + 3 x^{2} = 4

y^2(dy)/(dx)+2xy^3=6x,y(0)=1

y^{2} \frac{d y}{d x} + 2 x y^{3} = 6 x, y (0) = 1

(dy)/(dx)=3.2y^{2/3},x(0)=1,y(0)=13

\frac{d y}{d x} = 3.2 y^{\frac{2}{3}}, x (0) = 1, y (0) = 13

Frequently Asked Questions (FAQ)

What is the solution for y^'=y^2(9x+5) ?
The solution for y^'=y^2(9x+5) is y=-2/(9x^2+10x+c_{1)}