Question #f058d

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Answer 1 · 2016-11-29T16:36:39

$tan y = \frac{x^{2}}{2}$ $tan y =x^2/2$

This is a variable separable differential equation

$\frac{d y}{d x} = x {cos}^{2} y$ $(dy)/dx=x cos^2 y$

$\frac{d y}{{cos}^{2} y} = x \cdot d x$ $(dy)/cos^2 y=x* dx$

${sec}^{2} y d y = x \cdot d x$ $sec^2 y" "dy=x* dx$

$\int {sec}^{2} y d y = \int x \cdot d x$ $int sec^2 y" "dy=int x* dx$

$tan y = \frac{x^{2}}{2} + C$ $tan y =x^2/2+C$

using $y = 0$ $y=0$ when $x = 0$ $x=0$

$tan 0 = \frac{0^{2}}{2} + C$ $tan 0 =0^2/2+C$

$C = 0$ $C=0$

final answer

$tan y = \frac{x^{2}}{2} + 0$ $tan y =x^2/2+0$

$tan y = \frac{x^{2}}{2}$ $tan y =x^2/2$

God bless....I hope the explanation is useful.

Answer 2 · 2016-11-29T16:38:02

The variables can be separated naturally:

$\frac{d y}{d x} = x {cos}^{2} (y) \Rightarrow \frac{d y}{{cos}^{2} (y)} = x d x$ $(dy)/(dx)=x cos^2(y) => (dy)/cos^2(y)=xdx$

Integrate both sides:

$\int x d x = \int \frac{d y}{{cos}^{2} (y)}$ $int xdx = int (dy)/cos^2(y)$

$\frac{x^{2}}{2} + C = tan y$ $x^2/2 + C =tany$

For $x = 0, tan y ∣_{y = 0} = C \Rightarrow C = 0$ $x=0, tany|_(y=0) = C => C=0$

$y (x) = arctan (\frac{x^{2}}{2})$ $y(x)=arctan(x^2/2)$