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Popular >

(x+1)y^'+y=0

Solution

(x + 1) y^{^{'}} + y = 0

Solution

y = \frac{c _{1}}{x + 1}

Solution steps

(x + 1) y^{'} (x) + y = 0

Solve separable ODE:

y = \frac{c _{1}}{x + 1}

y = \frac{c _{1}}{x + 1}

Graph

Popular Examples

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

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

(dy)/(dx)=6xy^2

\frac{d y}{d x} = 6 x y^{2}

(dy)/(dx)=(8e^x)/(e^y)

\frac{d y}{d x} = \frac{8 e ^{x}}{e ^{y}}

(dy)/(dt)=ysqrt(t-1),y(1)=4

\frac{d y}{d t} = y t - 1, y (1) = 4

6y^'-3y-88=5y^',y(3)=300

6 y^{^{'}} - 3 y - 88 = 5 y^{^{'}}, y (3) = 300

Frequently Asked Questions (FAQ)

What is the solution for (x+1)y^'+y=0 ?
The solution for (x+1)y^'+y=0 is y=(c_{1})/(x+1)