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

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