Question

Use a substitution method to solve both of the following DEs, stating the general solution clearly, and showing all work clearly. a. dy/dx - y = e^xy^2 (Solve explicitly for y.) b. dy/dx = x + y/x - y (You can leave the General Solution in implicit form.)

Answer 1

$1.\rightarrow \frac{dy}{dx}-y=e^x y^2\\\\\rightarrow \frac{1}{y^2}\frac{dy}{dx}-\frac{1}{y}=e^x\\\\ \text{put},\frac{-1}{y}=z\\\\ \frac{dy}{y^2} =d z\\\\ \frac{dy}{dx} \times \frac{1}{y^2}=\frac{dz}{dx}\\\\\frac{dz}{dx} +z=e^x\\\\ \text{Integrating factor}=e^{\int {1} \, dx}\\\\=e^x \\\\ \text{Multiplying both sides by }e^x\\\\e^x(\frac{dz}{dx} +z)=e^{2x}\\\\ \text{Integrating both sides}\\\\z e^x=\frac{e^{2x}}{2}+C\\\\ \frac{-e^x}{y}=\frac{e^{2x}}{2}+C$

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$\rightarrow \frac{dy}{dx}=x+\frac{y}{x}-y\\\\\rightarrow \frac{dy}{dx}-x=\frac{y}{x}-y\\\\\rightarrow \frac{dy}{dx}+y(1-\frac{1}{x})=x\\\\\text{Integrating factor}=e^{\int{1-\frac{1}{x}}\,dx}\\\\=e^{x-\log x}\\\\ \text{Multiplying both sides by} e^{x-\log x}\\\\e^{x-\log x}\times[\frac{dy}{dx}+y(1-\frac{1}{x})]=x \times e^{x-\log x}\\\\y\times e^{x-\log x} =\int x \times e^{x-\log x} \, dx}\\\\y\times e^{x-\log x}=\int x \times \frac{e^x}{e^{\log x}}\,dx\\\\y\times e^{x-\log x}=\int x \times \frac{e^x}{x} \, dx\\\\y\times e^{x-\log x}=e^x+K$