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Popular >

(1+x^2)y^'+2xy=x

Solution

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

Solution

y = \frac{c _{1}}{2 ( x ^{2} + 1 )} + \frac{1}{2}

Solution steps

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

Solve separable ODE:

y = \frac{c _{1}}{2 ( x ^{2} + 1 )} + \frac{1}{2}

y = \frac{c _{1}}{2 ( x ^{2} + 1 )} + \frac{1}{2}

Popular Examples

y^'tan(x)=9a+y,y(pi/3)=9a

y^{^{'}} tan (x) = 9 a + y, y (\frac{π}{3}) = 9 a

x^{5} \cdot y^{^{'}} + y^{5} = 0

(dx)/(1+x^2)+(dy)/(1+y^2)=0

\frac{d x}{1 + x ^{2}} + \frac{d y}{1 + y ^{2}} = 0

xcos(y)(dy)/(dx)-sin(y)=0

x cos (y) \frac{d y}{d x} - sin (y) = 0

x^{(-2)}y^{(-5)}dx+x^{(-3)}y^{(-4)}dy=0

x^{(- 2)} y^{(- 5)} d x + x^{(- 3)} y^{(- 4)} d y = 0

Frequently Asked Questions (FAQ)

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