Let f be twice continuously differentiable. Consider ω = (x…

Let $f$ be twice continuously differentiable. Consider \[ \omega = \left(x\,\frac{\partial u}{\partial x}+y\right)dx + \left[x\,\frac{\partial u}{\partial y}+x-2xy+2yf'(u-x)\right]dy. \]
1° Form and integrate the PDE that $u=u(x,y)$ must satisfy for $\omega$ to be a total differential.
2° Find the integral satisfying $u=y^2$ when $x=0$. For the resulting $u(x,y)$, find the function whose total differential is $\omega$.

Step 1 · Exactness condition

With $P=xu_x+y$, $Q=xu_y+x-2xy+2yf'(u-x)$: \[ P_y = xu_{xy}+1, \qquad Q_x = u_y+xu_{xy}+1-2y+2yf''(u-x)(u_x-1). \] Setting $P_y=Q_x$ and cancelling $xu_{xy}+1$: \[ u_y - 2y + 2yf''(u-x)(u_x-1) = 0. \] Since this must hold for an arbitrary function $f$, the coefficient of $f''$ must vanish independently, forcing $u_x=1$; the remaining terms then require $u_y=2y$.

Step 2 · Integrate the system $u_x=1$, $u_y=2y$

From $u_x=1$: $u=x+\varphi(y)$. Substituting into $u_y=2y$: $\varphi'(y)=2y \Rightarrow \varphi(y)=y^2+C$. General solution: \[ u = x+y^2+C. \]

Step 3 · Apply the initial condition

At $x=0$: $u=y^2=0+y^2+C \Rightarrow C=0$, so $u=x+y^2$.

Step 4 · Find the function with total differential $\omega$

Substituting $u=x+y^2$ (so $u-x=y^2$) into $\omega$: \[ P=x+y, \qquad Q = x+2yf'(y^2), \] which is exact ($P_y=Q_x=1$). Integrating $F_x=P$: $F=\tfrac{x^2}{2}+xy+g(y)$; matching $F_y=Q$ gives $g'(y)=2yf'(y^2)$, so $g(y)=f(y^2)+C_0$.

Verification

With $u=x+y^2$: $u_x=1$, $u_y=2y$. $\checkmark$ At $x=0$: $u=y^2$. $\checkmark$ And $dF=(x+y)\,dx+(x+2yf'(y^2))\,dy=\omega$. $\checkmark$