Mathematics - Differential Equation Question with Solution | TestHub

put $\frac{x}{y}=t$ $\mathrm{x}=\mathrm{yt}$ $\frac{dx}{dy}=t+\frac{ydt}{dy}$ $\therefore \left(1+{\mathrm{e}}^{\mathrm{t}}\right)\left(\mathrm{t}+\frac{\mathrm{ydt}}{\mathrm{dy}}\right)+{\mathrm{e}}^{\mathrm{t}}(1-\mathrm{t})=0$ $t+\frac{ydt}{dy}=\frac{e^t(t-1)}{1+e^t}$ $\frac{ydt}{dy}=\frac{e^t(t-1)}{\left(e^t+1\right)}-t$ $\frac{ydt}{dy}=-\frac{e^t+t}{e^t+1}$ $\frac{\left(e^t+1\right)dt}{\left(e^t+t\right)}=-\frac{dy}{y}$ $\ln \left(e^t+t\right)=-\ln y+\ln C$ $\ln \left(e^t+t\right)+\ln y=\ln C$ $\left({\mathrm{e}}^{\mathrm{t}}+\mathrm{t}\right)\mathrm{y}=\mathrm{C}$ $x+y{e}^{x/y}=C$