Answer to Question #311635 in Differential Equations for Papi Chulo

Question #311635

Find a general solution of the following differential equation



dy/dx - 5y = (xe^-2x)*y^-2

1
Expert's answer
2022-03-15T17:26:31-0400

"\\text{Solve Bernoulli's equation } \\frac{d y(x)}{d x}-5 y(x)=\\frac{e^{-2 x} x}{y(x)^{2}} :\\\\[4mm]\n\\text{Multiply both sides by }3 y(x)^{2} :\\\\[2mm]\n3 \\frac{d y(x)}{d x} y(x)^{2}-15 y(x)^{3}=3 e^{-2 x} x\\\\[4mm]\n\\text{Let } v(x)=y(x)^{3},\\text{ which gives } \\frac{d v(x)}{d x}=3 y(x)^{2} \\frac{d y(x)}{d x}:\\\\\n\\frac{d v(x)}{d x}-15 v(x)=3 e^{-2 x} x\\\\[4mm]\n\\text{ Let } \\mu(x)=e^{\\int-15 d x}=e^{-15 x}.\\\\\n\\text{ Multiply both sides by } \\mu(x): \\\\[2mm]\ne^{-15 x} \\frac{d v(x)}{d x}-\\left(15 e^{-15 x}\\right) v(x)=3 e^{-17 x} x\\\\\n\\text{ Substitute } -15 e^{-15 x}=\\frac{d}{d x}\\left(e^{-15 x}\\right) :\\\\\ne^{-15 x} \\frac{d v(x)}{d x}+\\frac{d}{d x}\\left(e^{-15 x}\\right) v(x)=3 e^{-17 x} x\\\\[4mm]\n\\text{ Apply the reverse product rule } f \\frac{d g}{d x}+g \\frac{d f}{d x}=\\frac{d}{d x}(f g) \\text{ to the left-hand side: } \\frac{d}{d x}\\left(e^{-15 x} v(x)\\right)=3 e^{-17 x} x\\\\"


"\\text{ Evaluate the integrals:}\\\\[2mm]\ne^{-15 x} v(x)=3 e^{-17 x}\\left(-\\frac{x}{17}-\\frac{1}{289}\\right)+c_{1}, \\text{ where } c_{1} \\text{ is an arbitrary constant.}\\\\[3mm]\n\\text{ Divide both sides by } \\mu(x)=e^{-15 x} :\\\\[2mm]\nv(x)=\\frac{1}{289} e^{-2 x}\\left(-51 x+289 c_{1} e^{17 x}-3\\right)\\\\[4mm]\n\\text{ Solve for } y(x) \\text{ in } v(x)=y(x)^{3}:\\\\[2mm]\n\\begin{aligned}\n&y(x)=\\frac{e^{-(2 x) \/ 3} \\sqrt[3]{-51 x+289 c_{1} e^{17 x}-3}}{17^{2 \/ 3}} \\\\\n&\\text { or } y(x)=-\\frac{\\sqrt[3]{-1} e^{-(2 x) \/ 3} \\sqrt[3]{-51 x+289 c_{1} e^{17 x}-3}}{17^{2 \/ 3}} \\\\\n&\\text { or } y(x)=\\frac{(-1)^{2 \/ 3} e^{-(2 x) \/ 3} \\sqrt[3]{-51 x+289 c_{1} e^{17 x}-3}}{17^{2 \/ 3}}\n\\end{aligned}\n\\\\[4mm]\n\\text{ Simplifying the arbitrary constants we have:}\\\\[4mm]\n\\begin{aligned}\n&y(x)=\\frac{e^{-(2 x) \/ 3} \\sqrt[3]{-51 x+c_{1} e^{17 x}-3}}{17^{2 \/ 3}} \\\\\n&\\text { or } y(x)=-\\frac{\\sqrt[3]{-1} e^{-(2 x) \/ 3} \\sqrt[3]{-51 x+c_{1} e^{17 x}-3}}{17^{2 \/ 3}} \\\\\n&\\text { or } y(x)=\\frac{(-1)^{2 \/ 3} e^{-(2 x) \/ 3} \\sqrt[3]{-51 x+c_{1} e^{17 x}-3}}{17^{2 \/ 3}}\n\\end{aligned}"


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