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Popular >

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

Solution

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

Solution

y = e^{a r c t a n (x) + c_{1}} + 2

Solution steps

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

Solve separable ODE:

y = e^{a r c t a n (x) + c_{1}} + 2

y = e^{a r c t a n (x) + c_{1}} + 2

Popular Examples

(dy)/(dx)=((y-1)/(x+3))

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

(dy)/(dx)+2xy-x=0

\frac{d y}{d x} + 2 x y - x = 0

(1+x)(dy)/(dx)+y=2e^{2pi}

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

y=e^{cos(x^2)}y^'

y = e^{c o s (x^{2})} y^{^{'}}

y(1-x^2)dy-x(1-y^2)dx=0

y (1 - x^{2}) d y - x (1 - y^{2}) d x = 0

Frequently Asked Questions (FAQ)

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