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Question Number 114176 by bemath last updated on 17/Sep/20

$$\left(1\right)3x^2\ln \left(y\right)dx+\frac{x^3}{y}dy=0$$$$\left(2\right)(e^{2x}+4)y{}^{\prime }=y$$$$\left(3\right)dz=t(t^2+1).e^{2z}dt$$

Commented by bobhans last updated on 17/Sep/20

$$\left(2\right)\frac{dy}{y}=\frac{dx}{e^{2x}+4}\Rightarrow \int \frac{dy}{y}=\int \frac{dx}{e^{2x}+4}$$$$\Rightarrow \ln \left(y\right)=\int \frac{dx}{e^{2x}+4}$$$$[let{e}^x=2\tan \psi \to e^{x}dx=2\sec {}^2\psi d\psi ]$$$$\iff \ln \left(y\right)=\int \frac{2\sec {}^2\psi }{2\tan \psi .4\sec {}^2\psi }d\psi$$$$\iff \ln \left(y\right)=\frac{1}{4}\int \frac{d\left(\sin \psi \right)}{\sin \psi }$$$$\ln \left(y\right)=\frac{1}{4}\ln \left(\sin \psi \right)+c$$$$\ln \left(y\right)=\ln \left(C\sqrt[4]{\sin \psi }\right)\Rightarrow y=C\sqrt[4]{\frac{e^x}{4+e^{2x}}}.$$

Answered by bobhans last updated on 17/Sep/20

Answered by Dwaipayan Shikari last updated on 17/Sep/20

$$\int \frac{dz}{e^{2z}}=\int t(t^2+1)dt$$$$-\frac{1}{2}{e}^{-2z}=\frac{t^4}{4}+\frac{t^2}{2}+C$$$$e^{-2z}=-\frac{t^4}{2}-t^2+C_1$$$$-2z=log(C_1-\frac{t^4}{2}-t^2)$$$$z=-\frac{1}{2}log(C_1-\frac{t^4}{2}-t^2)$$$$z=log\left(\frac{1}{\sqrt{C_1-\frac{t^4}{2}-t^2}}\right)$$

Answered by mathmax by abdo last updated on 18/Sep/20

$$2)({\mathrm{e}}^{2\mathrm{x}}+4){\mathrm{y}}^{{}^{\prime }}=\mathrm{y}\Rightarrow \frac{{\mathrm{y}}^{{}^{\prime }}}{\mathrm{y}}=\frac{1}{{\mathrm{e}}^{2\mathrm{x}}+4}\Rightarrow \ln \mid \mathrm{y}\mid =\int \frac{\mathrm{dx}}{{\mathrm{e}}^{2\mathrm{x}}+4}$$$${=}_{{\mathrm{e}}^{\mathrm{x}}=\mathrm{t}}\int \frac{\mathrm{dt}}{\mathrm{t}({\mathrm{t}}^2+4)}=\frac{1}{4}\int (\frac{1}{\mathrm{t}}-\frac{\mathrm{t}}{{\mathrm{t}}^2+4})\mathrm{dt}=\frac{1}{4}\ln \left(\mathrm{t}\right)-\frac{1}{8}\ln ({\mathrm{t}}^2+4)+\mathrm{c}$$$$\Rightarrow \mathrm{y}\left(\mathrm{x}\right)=\mathrm{k}{\mathrm{t}}^{\frac{1}{4}}.{({\mathrm{t}}^2+4)}^{-\frac{1}{8}}$$$$$$

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