the general solution of x^2y+2xy=4x cos(2x)sqrt y

The given differential equation is:

\[
x^2y + 2xy = 4x \cos(2x) \sqrt{y}
\]

### Step 1: Simplify the equation
Let's rewrite the equation:

\[
xy(x + 2) = 4x \cos(2x) \sqrt{y}
\]

Now divide both sides by \(x\) (assuming \(x \neq 0\)):

\[
y(x + 2) = 4 \cos(2x) \sqrt{y}
\]

### Step 2: Isolate terms involving \(y\)
Now, divide both sides by \( \sqrt{y} \) (assuming \(y \neq 0\)):

\[
\sqrt{y}(x + 2) = 4 \cos(2x)
\]

### Step 3: Solve for \(y\)
Square both sides to eliminate the square root:

\[
y(x + 2)^2 = 16 \cos^2(2x)
\]

Finally, solve for \(y\):

\[
y = \frac{16 \cos^2(2x)}{(x + 2)^2}
\]

Thus, the general solution to the differential equation is:

\[
y = \frac{16 \cos^2(2x)}{(x + 2)^2}
\]

### Would you like further details or have any questions?

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Here are 5 related questions for further exploration:

1. How would the solution change if the right-hand side of the equation had a different trigonometric function?
2. What is the behavior of the solution as \(x \to \infty\)?
3. Can the solution be written in terms of a series expansion for small \(x\)?
4. How would you solve the equation if there were higher powers of \(y\) involved?
5. How does this type of equation relate to separable or exact differential equations?

**Tip:** Always check for simplifying assumptions, such as \(x \neq 0\) or \(y \neq 0\), when solving differential equations.

The general solution of x^2y + 2xy = 4x cos(2x) sqrt(y)

Learn how to solve the differential equation x^2y + 2xy = 4x cos(2x) sqrt(y) using algebraic manipulation and trigonometric identities. This detailed step-by-step guide walks through simplifying the equation, isolating terms, and solving for y. Suitable for undergraduate-level math students.

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