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Let's walk through the suggested solution for **Q1** from the assignment:

You are asked to verify whether the expression \( y^2 - 4x^2 + e^{xy} = C \) is a solution to the given ODE:  
\[ (8x - y e^{xy}) - (2y + x e^{xy})y' = 0. \]

### Suggested Solution Breakdown:

1. **Differentiate the given equation** \( y^2 - 4x^2 + e^{xy} = C \) with respect to \( x \):
   \[
   (y^2 - 4x^2 + e^{xy})' = (C)'.
   \]
   This gives:
   \[
   2yy' - 8x + (y + x y')e^{xy} = 0.
   \]
   Simplifying further, we get:
   \[
   2yy' + x y' e^{xy} - 8x + y e^{xy} = 0.
   \]

2. **Isolate \( y' \)** in the resulting equation:
   \[
   2yy' + x y' e^{xy} = 8x - y e^{xy}.
   \]
   Solving for \( y' \):
   \[
   y' = \frac{8x - y e^{xy}}{2y + x e^{xy}}.
   \]

3. **Substitute \( y' \)** into the original ODE \( (8x - y e^{xy}) - (2y + x e^{xy})y' = 0 \) to verify the solution:
   Substituting \( y' = \frac{8x - y e^{xy}}{2y + x e^{xy}} \) into the ODE, we get:
   \[
   (8x - y e^{xy}) - (2y + x e^{xy})\frac{8x - y e^{xy}}{2y + x e^{xy}} = 0.
   \]
   Simplifying the right-hand side:
   \[
   (8x - y e^{xy}) - (8x - y e^{xy}) = 0.
   \]
   Since this equation holds true, we confirm that \( y^2 - 4x^2 + e^{xy} = C \) is indeed a solution to the given ODE.

Would you like more details or have any questions about specific steps? 

### Related questions to explore further:
1. How do you systematically differentiate implicit functions in similar cases?
2. Why is solving for \( y' \) important when verifying a solution to an ODE?
3. Can this method be applied to non-linear differential equations in general?
4. What are other common techniques for verifying solutions to ODEs?
5. How does the presence of exponential terms affect solving differential equations?

### Tip: 
When differentiating, carefully handle exponential terms that involve products like \( e^{xy} \), applying both product and chain rules effectively.

This page explains how to verify that y^2 - 4x^2 + e^(xy) = C is a solution to a given ODE using implicit differentiation and solving for y'. It includes step-by-step calculations and is suitable for undergraduate students studying differential equations.

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