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Popular >

(x^2-9)y^'+xy=0

Solution

(x^{2} - 9) y^{^{'}} + x y = 0

Solution

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

Solution steps

(x^{2} - 9) y^{'} (x) + x y = 0

Solve separable ODE:

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

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

Popular Examples

y^'=(2y^2)/(x^2)

y^{^{'}} = \frac{2 y ^{2}}{x ^{2}}

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

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

(dr)/(dt)=0.1872*r

\frac{d r}{d t} = 0.1872 \cdot r

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

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

y^'=(x^2-4)(3y+2)

y^{^{'}} = (x^{2} - 4) (3 y + 2)

Frequently Asked Questions (FAQ)

What is the solution for (x^2-9)y^'+xy=0 ?
The solution for (x^2-9)y^'+xy=0 is y=(c_{1})/(sqrt(x^2-9))