Im solving initial value problem $$ \frac {dy}{dx} + xy = xy^2; y(0)=3$$ After applying Bernoulli's equation method i obtained $$ \frac {dv}{dx} -xv = -x$$ So, $$ p(x) = -x, q(x) = -x $$ For finding integrating factor $$u(x)=e^ {-\int xdx}=e^ {-\frac {x^2}{2}}$$ $$ y= \frac {\int u(x)q(x)dx+C}{u(x)}={\int u(x)q(x)dx}$$ So, $$-\int xe^ {-\frac {x^2}{2}}dx $$ Please help further or guide me if i did something wrong.
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This seems fine. the only point is $v=\frac1y$, so $y=\frac1v$ and $$u(x)v(x)-1\cdot\frac13=u(x)v(x)-u(0)v(0)=\int_0^x (u(t)v(t))'=\int_0^x u(t)q(t)dt=-\int_0^x te^{-\frac{t^2}{2}}dt$$ So $$v(x)=e^{\frac{x^2}{2}}\left(\frac13-\int_0^x te^{-\frac{t^2}{2}}dt\right),\hspace{10pt} y(x)=\frac{1}{v(x)}$$ |
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Why don't you try separation of variables? This leads to a closed form solution that exists for all $x < \sqrt{2 \log \frac{3}{2}}$. |
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