Question #223875

Solve dy/dx=4x^{2}(y-x)^{2}+y/x if y=x is a particular solution

Expert's answer

"\\frac{dy}{dx}= 4x\u00b2 (y-x)\u00b2\\\\\ndy= 4x\u00b2 (y-x)\u00b2dx\\\\\ndy- 4x\u00b2 (y-x)\u00b2dx=0\\\\"

We know that "\\frac{\\partial F}{\\partial x}=\\left(4x^2(y-x)^2\\right)"

"\\implies \\int\\frac{\\partial F }{\\partial x}=\\int\\left(4x^2(y-x)^2\\right)\\\\\n\\mathrm{Take\\:the\\:constant\\:out}:\\quad \\int a\\cdot f\\left(x\\right)dx=a\\cdot \\int f\\left(x\\right)dx\\\\\n=4\\cdot \\int \\left(y-x\\right)^2x^2dx\\\\\n4\\cdot \\int \\:y^2x^2-2yx^3+x^4dx\\\\\n\\mathrm{Apply\\:the\\:Sum\\:Rule}:\\quad \\int f\\left(x\\right)\\pm g\\left(x\\right)dx=\\int f\\left(x\\right)dx\\pm \\int g\\left(x\\right)dx\\\\\n=4\\left(\\int \\:y^2x^2dx-\\int \\:2yx^3dx+\\int \\:x^4dx\\right)\\\\\n=4\\left(\\frac{y^2x^3}{3}-\\frac{yx^4}{2}+\\frac{x^5}{5}\\right)+f'(y)\\\\\n\\frac{d}{dy}\\left(4\\left(\\frac{y^2x^3}{3}-\\frac{yx^4}{2}+\\frac{x^5}{5}\\right)\\right)+ f'(y)\\\\\n=4\\frac{d}{dy}\\left(\\frac{y^2x^3}{3}-\\frac{yx^4}{2}+\\frac{x^5}{5}\\right)+\\int f'(y)\\\\\\\\\n=4\\left(\\frac{d}{dy}\\left(\\frac{y^2x^3}{3}\\right)-\\frac{d}{dy}\\left(\\frac{yx^4}{2}\\right)+\\frac{d}{dy}\\left(\\frac{x^5}{5}\\right)\\right)+\\int f'(y)\\\\\\\\\n=4\\left(-\\frac{x^4}{2}+\\frac{2yx^3}{3}\\right)+\\int f(y)\\\\\nF= 4\\left(-\\frac{x^4}{2}+\\frac{2yx^3}{3}\\right)+y\\\\"

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