Answer to Question #229250 in Chemical Engineering for Lokika

Question

Solve dy/dx= 4x² (y-x)² + y/x if y=x is a particular solution.

Answer 1

$\frac{dy}{dx}= 4x² (y-x)²\\ dy= 4x² (y-x)²dx\\ dy- 4x² (y-x)²dx=0\\$We know that $\frac{\partial F}{\partial x}=\left(4x^2(y-x)^2\right)$

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