Find the general solution of 2y^4 p - xy q = x√(z^2+1), p =…
Find the general solution of \[ 2y^4\,p - xy\,q = x\sqrt{z^2+1}, \qquad p = \frac{\partial z}{\partial x},\quad q = \frac{\partial z}{\partial y}. \]
Explanation of Method
A Lagrange PDE with $P=2y^4$, $Q=-xy$, $R=x\sqrt{z^2+1}$. The first integral is separable directly; the second follows by substituting $y^4=c_1-x^2$ (from the first integral) into the first/third ratio, giving a separable equation in $z$ and $x$.
Step 1 · Characteristic equations
\[ \frac{dx}{2y^4} = \frac{dy}{-xy} = \frac{dz}{x\sqrt{z^2+1}}. \tag{1} \]
Step 2 · First integral
From $\dfrac{dx}{2y^4}=\dfrac{dy}{-xy}$: $-xy\,dx=2y^4\,dy \Rightarrow x\,dx=-2y^3\,dy$. Integrating: $\dfrac{x^2}{2}=-\dfrac{y^4}{2}+\text{const}$, so \[ x^2 + y^4 = c_1. \tag{I} \]
Step 3 · Second integral
From $\dfrac{dx}{2y^4}=\dfrac{dz}{x\sqrt{z^2+1}}$: $x\sqrt{z^2+1}\,dx=2y^4\,dz$. Using $y^4=c_1-x^2$ from (I): \[ \frac{dz}{\sqrt{z^2+1}} = \frac{x\,dx}{2(c_1-x^2)}. \] The left side integrates to $\sinh^{-1}z=\ln(z+\sqrt{z^2+1})$; with $u=c_1-x^2$, the right side integrates to $-\tfrac14\ln(c_1-x^2)=-\tfrac14\ln(y^4)=-\ln y$. Hence \[ \ln\!\left(z+\sqrt{z^2+1}\right) = -\ln y + \text{const} \;\Longrightarrow\; y\!\left(z+\sqrt{z^2+1}\right) = c_2. \tag{II} \]
Step 4 · General solution
The general solution is $c_2=\Phi(c_1)$: \[ y\!\left(z + \sqrt{z^2+1}\right) = \Phi\!\left(x^2+y^4\right), \] where $\Phi$ is an arbitrary differentiable function.
Verification
First integral: $P c_{1x}+Qc_{1y}=2y^4\cdot2x+(-xy)\cdot4y^3
=4xy^4-4xy^4=0. \checkmark$
PDE check: differentiating $y(z+\sqrt{z^2+1})=\Phi(x^2+y^4)$ with respect
to $x$ and $y$ and eliminating $\Phi'$ between the two relations (after factoring the
common term $(z+\sqrt{z^2+1})/\sqrt{z^2+1}$) yields
\[ 2y^4 p - xyq = x\sqrt{z^2+1}. \checkmark \]
Result
The general solution is: