anonymous 3 years ago Diff eqn dy/dx= (ycos(x))/(1+2y^2) int condt y(0)=1

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1. hartnn

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

2. hartnn

can you ? or need help with that ?

3. Mimi_x3

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..

4. anonymous

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

5. Mimi_x3

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