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Popular >

y^'=5y^2ln(x)

Solution

y^{^{'}} = 5 y^{2} ln (x)

Solution

y = - \frac{1}{5 x ln ( x ) - 5 x + c _{1}}

Solution steps

y^{'} (x) = 5 y^{2} ln (x)

Solve separable ODE:

y = - \frac{1}{5 x ln ( x ) - 5 x + c _{1}}

y = - \frac{1}{5 x ln ( x ) - 5 x + c _{1}}

Popular Examples

(t^2-16t+15)y^'=y

(t^{2} - 16 t + 15) y^{^{'}} = y

y^'=(1-2x)y^2,y(0)=(-1)/6

y^{^{'}} = (1 - 2 x) y^{2}, y (0) = \frac{- 1}{6}

(dy)/(dx)-y^5=0,y(0)=1

\frac{d y}{d x} - y^{5} = 0, y (0) = 1

2(x^2+1)(dy)/(dx)=x(4-y^2)

2 (x^{2} + 1) \frac{d y}{d x} = x (4 - y^{2})

(1+x^2)(dy)/(dx)+2xy=-x,x>0

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

Frequently Asked Questions (FAQ)

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