The image you provided contains two equations. Let's break them down:

1. $y' + 4xy = 2xe^{-x^2}$
2. $\sqrt{y}, z = y^{1-2}$

1. First Equation:

This is a first-order linear differential equation: $y' + 4xy = 2xe^{-x^2}$

We can solve this using an integrating factor. The integrating factor $\mu(x)$ for a differential equation of the form: $y' + P(x)y = Q(x)$ is given by: $\mu(x) = e^{\int P(x) dx}$

Here, $P(x) = 4x$, so the integrating factor is: $\mu(x) = e^{\int 4x dx} = e^{2x^2}$

Multiplying both sides of the original equation by this integrating factor, we get: $e^{2x^2}y' + 4xe^{2x^2}y = 2xe^{2x^2 - x^2}$

Simplifying the right-hand side: $e^{2x^2 - x^2} = e^{x^2}$ Thus, the equation becomes: $e^{2x^2}y' + 4xe^{2x^2}y = 2xe^{x^2}$

This can be solved further by recognizing the left-hand side as the derivative of $e^{2x^2}y$.

2. Second Equation:

The second part of the equation appears to define two variables: $\sqrt{y}, z = y^{1-2}$

This equation suggests:

• $\sqrt{y} = y^{1/2}$
• $z = y^{1-2} = y^{-1}$

Thus, $z = \frac{1}{y}$.

Would you like me to proceed with a full solution for the differential equation or further details on the second equation? Let me know if you need clarification!

Here are 5 related questions to further explore:

1. How do you solve first-order linear differential equations in general?
2. What is the purpose of the integrating factor in differential equations?
3. How do you find general solutions to linear differential equations?
4. What is the interpretation of $z = y^{-1}$ in the context of this system?
5. How can you verify the solution to a differential equation?

Tip: When solving linear differential equations, always look for patterns like the integrating factor, which simplifies the process of finding the solution.