Question

Diff eqn dy/dx= (ycos(x))/(1+2y^2) int condt y(0)=1

Answer 1

this is easily separable.
bring it in the form f(x)dx = f(y)dy
then integrate both sides.

Answer 2

can you ? or need help with that ?

Answer 3

because im bored...\[\frac{dy}{dx} = \frac{ycos(x)}{1+2y^{2}} \]
\[=>\frac{dy}{dx} = \frac{y}{1+2y^{2}} *\cos(x) => dy = \frac{y}{1+2y^{2}} *\cos(x)dx => \frac{dy(1+2y^{2})}{y} = cos(x) \]
\[=> \int\limits\frac{1+2y^{2}}{y} dy = \int\limits \cos(x) \]
\[\int\limits\frac{1}{y} dy + \frac{2y^{2}}{y} dy = \ln(y) + \frac{y^{2}}{2} \]
\[=> \ln(y)+\frac{y^{2}}{2} = sinx +c\]
very easy DE..

Answer 4

hmm, it's \[ \ln y+y^2=\sin x + C\]
now put \(x=0\) and \(y=1\) to find \(C\)

Answer 5

wait there's a typo..\[ \int\limits\frac{2y^{2}}{y} dy => \int\limits2y dy => y^2\]