Answer to Question #224869 in Chemical Engineering for Lokika

Question

Solve the initial value problem y²y'-y^3 tanx = sin x cos² x, y(0)=1

Answer 1

$y^2y'\:-y^3\tan \left(x\right)=\sin \left(x\right)\cos ^2\left(x\right)\\ \mathrm{A\:first\:order\:Bernoulli\:ODE\:has\:the\:form\:of\:}y'+p\left(x\right)y=q\left(x\right)y^n\\ y'\:-\tan \left(x\right)y=\sin \left(x\right)\cos ^2\left(x\right)y^{-2}\\ \mathrm{The\:general\:solution\:is\:obtained\:by\:substituting\:}v=y^{1-n}\mathrm{\:and\:solving\:}\frac{1}{1-n}v'+p\left(x\right)v=q\left(x\right)\\ ν=-\frac{\cos ^3\left(x\right)}{2}+\frac{c_1}{\cos ^3\left(x\right)}\\ y^3=-\frac{\cos ^3\left(x\right)}{2}+\frac{c_1}{\cos ^3\left(x\right)}\\ y=\frac{2^{\frac{2}{3}}\sqrt[3]{-\cos ^6\left(x\right)+3}}{2\cos \left(x\right)}\\ y=\frac{2^{\frac{2}{3}}\sqrt[3]{-\cos ^6\left(x\right)+3}}{2\cos \left(x\right)}$