Verifying a Solution for a First-Order Partial Differential Equation

Posted on Oct 3, 2025 in Mathematical Engineering

Problem Statement and Given Solution

We are tasked with showing that the given function $u$ satisfies the following partial differential equation (PDE):

𝑥𝜕𝑢𝜕𝑥−𝑦𝜕𝑢𝜕𝑦=𝑦2𝑢3

The proposed solution is:

𝑢=(1+2𝑥𝑦+𝑦2)−12

1. Calculate the Partial Derivative 𝜕𝑢/𝜕𝑥

Using the chain rule, we differentiate $u$ with respect to $x$:

𝜕𝑢𝜕𝑥=−12(1+2𝑥𝑦+𝑦2)−32⋅(2𝑦)=−𝑦(1+2𝑥𝑦+𝑦2)−32

2. Calculate the Partial Derivative 𝜕𝑢/𝜕𝑦

Using the chain rule, we differentiate $u$ with respect to $y$:

𝜕𝑢𝜕𝑦=−12(1+2𝑥𝑦+𝑦2)−32⋅(2𝑥+2𝑦)=−(𝑥+𝑦)(1+2𝑥𝑦+𝑦2)−32

3. Substitute Derivatives into the Left-Hand Side (LHS)

We substitute the calculated partial derivatives into the expression $𝑥𝜕𝑢𝜕𝑥−𝑦𝜕𝑢𝜕𝑦$:

𝑥[−𝑦(1+2𝑥𝑦+𝑦2)−32]−𝑦[−(𝑥+𝑦)(1+2𝑥𝑦+𝑦2)−32]

Simplifying the signs:

=−𝑥𝑦(1+2𝑥𝑦+𝑦2)−32+𝑦(𝑥+𝑦)(1+2𝑥𝑦+𝑦2)−32

4. Factor and Simplify the LHS

We factor out the common term $(1+2𝑥𝑦+𝑦2)−32$:

=(1+2𝑥𝑦+𝑦2)−32[−𝑥𝑦+𝑦(𝑥+𝑦)]

Expand the terms inside the brackets:

=(1+2𝑥𝑦+𝑦2)−32[−𝑥𝑦+𝑥𝑦+𝑦2]

The $xy$ terms cancel out, leaving the simplified LHS:

=(1+2𝑥𝑦+𝑦2)−32[𝑦2]

5. Express the Result in Terms of $u$

We relate the simplified LHS back to the original function $u$. Since $𝑢=(1+2𝑥𝑦+𝑦2)−12$, we find $u^3$:

𝑢3=((1+2𝑥𝑦+𝑦2)−12)3=(1+2𝑥𝑦+𝑦2)−32

Substituting $u^3$ into the final LHS expression from Step 4:

𝑦2(1+2𝑥𝑦+𝑦2)−32=𝑦2𝑢3

Thus, we have successfully shown that $𝑥𝜕𝑢𝜕𝑥−𝑦𝜕𝑢𝜕𝑦=𝑦2𝑢3$. The given solution satisfies the PDE.