Find the general solution of the given differential equation: x dy/dx + y = x² sin x

To solve the differential equation given by x dy/dx + y = x² sin x, we first rewrite it in a more standard form. We can rearrange the equation as follows:

dy/dx = (x² sin x – y)/x.

This is a first-order linear differential equation. We can recognize it as dy/dx + (1/x)y = x sin x, which fits the standard form of a linear differential equation dy/dx + P(x)y = Q(x), with:

  • P(x) = 1/x
  • Q(x) = x sin x

Next, we need to find the integrating factor, which is given by:

μ(x) = e^{∫P(x) dx} = e^{∫1/x dx} = e^{ln|x|} = |x|.

For simplicity, we can consider μ(x) = x since we are working with x > 0.

We now multiply the entire differential equation by the integrating factor:

x dy/dx + y = x² sin x.

The left-hand side can be rewritten as the derivative of a product:

d/dx (xy) = x² sin x.

Next, we integrate both sides with respect to x:

∫d(xy) = ∫x² sin x dx.

To solve the right side, we can use integration by parts. Let:

  • u = x², hence du = 2x dx
  • dv = sin x dx, hence v = -cos x

Applying integration by parts:

∫x² sin x dx = -x² cos x – ∫(-cos x)(2x) dx.

This simplifies to:

∫x² sin x dx = -x² cos x + 2∫x cos x dx.

Again using integration by parts on ∫x cos x dx:

  • u = x, hence du = dx
  • dv = cos x dx, hence v = sin x

Thus, we find:

∫x cos x dx = x sin x – ∫sin x dx = x sin x + cos x.

Now substituting this back in:

∫x² sin x dx = -x² cos x + 2(x sin x + cos x).

Putting it all together, we have:

xy = -x² cos x + 2x sin x + 2cos x + C,

where C is the constant of integration.

Finally, we can isolate y:

y = -x cos x + 2 sin x + (2/x) cos x + C/x.

This gives us the general solution of the original differential equation:

y = -x cos x + 2 sin x + (2/x) cos x + C/x.

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