Question

Solve the differential equation y′=x(1+y2)
y=tan(x/2+c)
y=−tan(x/2+c)
y=−tan(x^2/2+c)
y=tan(x^2/2+c)

Answer 1

Answer on Question #79196 – Math – Differential Equations

Question

Solve the differential equation $y' = x(1 + y^2)$

$$y = \tan \left(\frac {x}{2} + c\right)$$ $$y = - \tan \left(\frac {x}{2} + c\right)$$ $$y = - \tan \left(\frac {x ^ {2}}{2} + c\right)$$ $$y = \tan \left(\frac {x ^ {2}}{2} + c\right)$$

Solution

$$\begin{array}{l} y ^ {\prime} = x (1 + y ^ {2}) \\ \frac {d y}{d x} = x (1 + y ^ {2}) \\ \frac {d y}{1 + y ^ {2}} = x d x \\ \int \frac {1}{1 + y ^ {2}} d y = \int x d x \\ \arctan y = \frac {x ^ {2}}{2} + c \\ y = \tan \left(\frac {x ^ {2}}{2} + c\right) \\ \end{array}$$

Answer: the fourth option is correct, $y = \tan \left(\frac{x^2}{2} + c\right)$.

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