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Popular >

xy^'+y=y^2,y(1)=-15

Solution

x y^{^{'}} + y = y^{2}, y (1) = - 15

Solution

y = \frac{15}{- 16 x + 15}

Solution steps

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

Solve Bernoulli ODE:

y = \frac{15}{- 16 x + 15}

y = \frac{15}{- 16 x + 15}

Popular Examples

area y=28-x^2,y=3x,y= 3/5 x area

y = 28 - x^{2}, y = 3 x, y = \frac{3}{5} x

integral of (16/2 x^3+12x^5-2x+1)

\int (\frac{16}{2} x^{3} + 12 x^{5} - 2 x + 1) d x

derivative of f(x)=5e^x(cos(x)-sin(x))derivative of

f (x) = 5 e^{x} (cos (x) - sin (x))

(\partial)/(\partial x)((x^2+y^2)^{0.5})

\frac{\partial}{\partial x} ((x^{2} + y^{2})^{0.5})

y^{^{''}} = 1

Frequently Asked Questions (FAQ)

What is the solution for xy^'+y=y^2,y(1)=-15 ?
The solution for xy^'+y=y^2,y(1)=-15 is y=(15)/(-16x+15)