I am given the initial value problem
\begin{array}{l} y' = \dfrac{7 x\,y}{7 x^{2}+2 y^{2}} \\ y(1)=1 \end{array}
where I must answer in the form of $F(x,y)=\frac{7}{4}$.
Here, I am also asked to use the substitution $y=xu$ to transform this differential equation into a separable differential equation in $u$. I am not sure how to go about doing this since I am not too familiar with differential equations. Any help would be greatly appreciated!
On
$$y=ux$$ $$dy=(u+xu')dx$$ $$y'=u+xu'$$ $$u+xu'=\frac{7ux^2}{7x^2+2u^2x^2}$$ $$u+xu'=\frac{7u}{7+2u^2}$$ $$(7+2u^2)(u+x\frac{du}{dx})=7u$$ $$7u+7x\frac{du}{dx}+2u^3+2u^2x\frac{du}{dx}=7u$$ $$x\frac{du}{dx}(7+2u^2)=-2u^3$$ $$\frac{1}{x} dx=-\frac{7}{2u^3}-\frac{1}{u}du$$ Then you can integrate both sides to find an equation in terms of $x,u$ $$\ln x=\frac{7}{4}u^{-2}-\ln u+C$$ Then use this $u=\frac{y}{x}$ to rewrite it and substitute values in to determine the value of the constant $C$.
When $x=1,y=1$ so when $x=1,u=1$
Thus $C=-\frac{7}{4}$
$$\ln y=\frac{7}{4}(\frac{x^2}{y^2}-1)$$
$$f(x,y)=\frac{y^2\ln y}{x^2-y^2}=\frac{7}{4}$$
Write $y'=u+u'x$ and substitute in you equation, then $$u+u'x=\dfrac{7x(ux)}{7x^2+2x^2u^2}=\dfrac{7u}{7+2u^2}$$ and $$\dfrac{7+2u^2}{2u^3}du=-\dfrac{dx}{x}$$ $$\dfrac{7}{2u^3}du+\dfrac{1}{u}du=-\dfrac{dx}{x}$$ $$\dfrac{7}{-4u^2}+\ln u=-\ln x+ C$$ $$\dfrac{7x^2}{-4y^2}+\ln y-\ln x=-\ln x+C$$ $$\ln y=C+\dfrac{7x^2}{4y^2}$$ set $y(1)=1$ so $C=-\dfrac74$ and $\color{blue}{-\ln y+\dfrac{7x^2}{4y^2}=\dfrac74}$.