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Question Number 79966 by M±th+et£s last updated on 29/Jan/20

Commented by M±th+et£s last updated on 29/Jan/20

$$solvethdODE$$

Answered by mr W last updated on 29/Jan/20

$$\frac{dx}{dy}=\frac{y^2+y^2{e}^{{\left(\frac{x}{y}\right)}^2}+2x^2{\left(\frac{x}{y}\right)}^2}{2{xye}^{{\left(\frac{x}{y}\right)}^2}}$$$$\frac{dx}{dy}=\frac{1+e^{{\left(\frac{x}{y}\right)}^2}+2{\left(\frac{x}{y}\right)}^4}{2\left(\frac{x}{y}\right)e^{{\left(\frac{x}{y}\right)}^2}}$$$$letu=\frac{x}{y},i.e.x=uy$$$$\frac{dx}{dy}=u+y\frac{du}{dy}$$$$u+y\frac{du}{dy}=\frac{1+e^{u^2}+2u^4}{2{ue}^{u^2}}$$$$y\frac{du}{dy}=\frac{1+e^{u^2}+2u^4-2u^2{e}^{u^2}}{2{ue}^{u^2}}$$$$\frac{2{ue}^{u^2}}{1+e^{u^2}+2u^4-2u^2{e}^{u^2}}du=\frac{dy}{y}$$$$\frac{e^{u^2}}{1+e^{u^2}+2u^4-2u^2{e}^{u^2}}{du}^2=\frac{dy}{y}$$$$lett=u^2$$$$\frac{e^t}{1+e^t+2t^2-2{te}^t}dt=\frac{dy}{y}$$$$\int \frac{e^t}{1+e^t+2t^2-2{te}^t}dt=\int \frac{dy}{y}$$$$\int \frac{e^t}{1+e^t+2t^2-2{te}^t}dt=\ln \left(cy\right)$$$$.....notintegrable...$$

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