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Popular >

y^'+(4y)/x =-4/x

Solution

y^{^{'}} + \frac{4 y}{x} = - \frac{4}{x}

Solution

y = \frac{c _{1}}{x ^{4}} - 1

Solution steps

y^{'} (x) + \frac{4 y}{x} = - \frac{4}{x}

Solve separable ODE:

y = \frac{c _{1}}{x ^{4}} - 1

y = \frac{c _{1}}{x ^{4}} - 1

Popular Examples

tangent of f(x)=(6x)/(x^2-3),\at x=3 tangent of

f (x) = \frac{6 x}{x ^{2} - 3}, at x = 3

(\partial)/(\partial z)(ye^{xz})

\frac{\partial}{\partial z} (y e^{x z})

9y^{''}-6y^'+8y=0

9 y^{^{''}} - 6 y^{^{'}} + 8 y = 0

(dy)/(dx)=3x^2(y-1)^3

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

derivative of 2(x-2^2)

\frac{d}{d x} (2 (x - 2)^{2})

Frequently Asked Questions (FAQ)

What is the solution for y^'+(4y)/x =-4/x ?
The solution for y^'+(4y)/x =-4/x is y=(c_{1})/(x^4)-1