Differential Equations Wizardry 🎩

Step 1: Identify the Type
This is a homogeneous differential equation because all terms are of the same degree.

Step 2: Substitute y = vx
Let y = vx ⇒ dy = v dx + x dv
Now rewrite the equation: [x + vx cos(v)] dx = x cos(v) (v dx + x dv)

Step 3: Simplify the Equation
(1 + v cos v) dx = v cos v dx + x cos v dv
⇒ dx = x cos v dv

Step 4: Separate Variables
(1/x) dx = cos v dv

Step 5: Integrate Both Sides
∫(1/x) dx = ∫cos v dv
⇒ ln|x| = sin v + C

Step 6: Substitute Back v = y/x
ln|x| = sin(y/x) + C

Step 1: Rewrite the Equation
(x³ + y³) dy = x²y dx ⇒ dy/dx = x²y/(x³ + y³)

Step 2: Substitute y = vx
Let y = vx ⇒ dy/dx = v + x dv/dx
Substitute into the equation: v + x dv/dx = x²(vx)/(x³ + (vx)³) = v/(1 + v³)

Step 3: Separate Variables
x dv/dx = v/(1 + v³) - v = -v⁴/(1 + v³)
⇒ (1 + v³)/v⁴ dv = -dx/x

Step 4: Integrate Both Sides
∫(1/v⁴ + 1/v) dv = -∫(1/x) dx
⇒ -1/(3v³) + ln|v| = -ln|x| + C

Step 5: Substitute Back v = y/x
-x³/(3y³) + ln|y/x| = -ln|x| + C
Simplify: -x³/(3y³) + ln|y| = C

Step 1: Rewrite the Equation
y eˣ dx - x eˣ dy = y dy

Step 2: Divide by y²
(eˣ dx)/y - (x eˣ dy)/y² = dy/y

Step 3: Recognize Exact Differential
Notice that d(x eˣ/y) = eˣ/y dx + x eˣ/y dx - x eˣ/y² dy

Step 4: Integrate
d(x eˣ/y) = dy/y ⇒ x eˣ/y = ln|y| + C

Step 1: Check for Exactness
M = 2xy, N = x² + 2y²
∂M/∂y = 2x, ∂N/∂x = 2x
Since ∂M/∂y = ∂N/∂x, the equation is exact.

Step 2: Find Potential Function
∫M dx = ∫2xy dx = x²y + h(y)
Differentiate with respect to y:
∂/∂y(x²y + h(y)) = x² + h'(y) = N = x² + 2y²
⇒ h'(y) = 2y² ⇒ h(y) = (2/3)y³ + C

Step 3: Write General Solution
The solution is x²y + (2/3)y³ = C

Step 1: Rewrite the Equation
dy/dx = (y² - 2xy)/(x² - 2xy)

Step 2: Substitute y = vx
Let y = vx ⇒ dy/dx = v + x dv/dx
Substitute into equation:
v + x dv/dx = (v²x² - 2x(vx))/(x² - 2x(vx)) = (v² - 2v)/(1 - 2v)

Step 3: Separate Variables
x dv/dx = (v² - 2v)/(1 - 2v) - v = (v² - 2v - v + 2v²)/(1 - 2v) = (3v² - 3v)/(1 - 2v)
⇒ (1 - 2v)/(3v² - 3v) dv = dx/x

Step 4: Integrate Both Sides
∫(1 - 2v)/(3v(v - 1)) dv = ∫(1/x) dx
Use partial fractions to solve the left integral.

Step 5: Final Solution
After integration and substitution back, we get:
(y - x)³ = C x y²

Step 1: Rewrite the Equation
dy/dx = y/x - cos²(y/x)

Step 2: Substitute y = vx
Let y = vx ⇒ dy/dx = v + x dv/dx
Substitute into equation:
v + x dv/dx = v - cos²(v) ⇒ x dv/dx = -cos²(v)

Step 3: Separate Variables
sec²(v) dv = -dx/x

Step 4: Integrate Both Sides
∫sec²(v) dv = -∫(1/x) dx ⇒ tan(v) = -ln|x| + C

Step 5: Substitute Back v = y/x
tan(y/x) = -ln|x| + C

Step 1: Rewrite the Equation
(1 + 3eˣ) dy = 3eˣ (y/x - 1) dx ⇒ dy/dx = 3eˣ (y/x - 1)/(1 + 3eˣ)

Step 2: Linear Form
dy/dx - [3eˣ/(x(1 + 3eˣ))] y = -3eˣ/(1 + 3eˣ)

Step 3: Integrating Factor
μ(x) = exp(∫-3eˣ/(x(1 + 3eˣ)) dx) = exp(-ln|x(1 + 3eˣ)|) = 1/(x(1 + 3eˣ))

Step 4: Solve
After applying the integrating factor and using the initial condition y(1) = 0, we get:
y = x [1 - (1 + 3e)/(1 + 3eˣ)]

Step 1: Rewrite the Equation
dy/dx = xy/(x² + y²)

Step 2: Substitute y = vx
Let y = vx ⇒ dy/dx = v + x dv/dx
Substitute into equation:
v + x dv/dx = x(vx)/(x² + (vx)²) = v/(1 + v²)

Step 3: Separate Variables
x dv/dx = v/(1 + v²) - v = -v³/(1 + v²)
⇒ (1 + v²)/v³ dv = -dx/x

Step 4: Integrate Both Sides
∫(1/v³ + 1/v) dv = -∫(1/x) dx ⇒ -1/(2v²) + ln|v| = -ln|x| + C

Step 5: Apply Initial Conditions
Using y(1) = 1 ⇒ v = 1 when x = 1:
-1/2 + 0 = 0 + C ⇒ C = -1/2
Final solution: y² = x²/(1 - 2ln|y/x|)
Find x₀ when y = e: x₀ = e^(3/2)