Question

find the solution of the differential equation that satisfies the given initial condition.

\\\frac{dy}{dx} = \frac{x \sin(x)}{y}, \quad y(0) = -9\\

Explanation:

⚡ Using what you learned: separable differential equations · 🆕 New: Integration by Parts
Using integration by parts to evaluate the x-interval integral

Step 1: Separate variables

Separate the variables \( y \) and \( x \):

$$ y \, dy = x \sin(x) \, dx $$

Integrate both sides:

$$ \int y \, dy = \int x \sin(x) \, dx $$

Step 2: Integrate both sides

The left side is straightforward:

$$ \int y \, dy = \frac{1}{2}y^2 $$

For the right side, use integration by parts, \(\int u \, dv = uv - \int v \, du\):

Let \( u = x \) and \( dv = \sin(x) \, dx \).
Then \( du = dx \) and \( v = -\cos(x) \).

$$ \int x \sin(x) \, dx = -x \cos(x) - \int (-\cos(x)) \, dx $$

$$ \int x \sin(x) \, dx = -x \cos(x) + \sin(x) + C $$

Equating both sides:

$$ \frac{1}{2}y^2 = -x \cos(x) + \sin(x) + C $$

Step 3: Apply the initial condition

Use \( y(0) = -9 \):

$$ \frac{1}{2}(-9)^2 = -0 \cdot \cos(0) + \sin(0) + C $$

$$ \frac{81}{2} = C $$

Substitute \( C \) back into the equation:

$$ \frac{1}{2}y^2 = -x \cos(x) + \sin(x) + \frac{81}{2} $$

Multiply the entire equation by 2:

$$ y^2 = -2x \cos(x) + 2\sin(x) + 81 $$

Step 4: Solve for y

Taking the square root gives two possible branches:

$$ y = \pm \sqrt{81 - 2x \cos(x) + 2\sin(x)} $$

Since the initial condition is \( y(0) = -9 \) (which is negative), we must choose the negative branch:

$$ y = -\sqrt{81 - 2x \cos(x) + 2\sin(x)} $$

Answer:

$$ y = -\sqrt{81 - 2x \cos(x) + 2\sin(x)} $$