Question

solve the differential equation $(x + 3)y + 3y = \frac{6\sin x}{(x + 3)^{2}},x>-3$.
the solution is $y = \square$.

Explanation:

Step1: Rewrite the differential equation in standard form

Divide the given equation \((x + 3)y'+3y=\frac{6\sin x}{(x + 3)^2}\) by \((x + 3)\) to get \(y'+\frac{3}{x + 3}y=\frac{6\sin x}{(x + 3)^3}\).
Here, \(P(x)=\frac{3}{x + 3}\) and \(Q(x)=\frac{6\sin x}{(x + 3)^3}\).

Step2: Find the integrating factor

The integrating factor \(\mu(x)=e^{\int P(x)dx}\).
Calculate \(\int\frac{3}{x + 3}dx=3\ln|x + 3|=\ln((x + 3)^3)\) (since \(x>-3\), we can drop the absolute - value), so \(\mu(x)=(x + 3)^3\).

Step3: Multiply the standard - form equation by the integrating factor

Multiply \(y'+\frac{3}{x + 3}y=\frac{6\sin x}{(x + 3)^3}\) by \((x + 3)^3\).
We get \((x + 3)^3y'+3(x + 3)^2y = 6\sin x\).
Notice that the left - hand side is the derivative of \((x + 3)^3y\) using the product rule \((uv)'=u'v+uv'\) (where \(u=(x + 3)^3\) and \(v = y\)).

Step4: Integrate both sides

Integrate \(\frac{d}{dx}[(x + 3)^3y]=6\sin x\) with respect to \(x\).
\(\int\frac{d}{dx}[(x + 3)^3y]dx=\int6\sin xdx\).
We know that \(\int6\sin xdx=-6\cos x + C\).
So \((x + 3)^3y=-6\cos x + C\).

Step5: Solve for \(y\)

Divide both sides of \((x + 3)^3y=-6\cos x + C\) by \((x + 3)^3\).
\(y=\frac{-6\cos x + C}{(x + 3)^3}\).

Answer:

\(y=\frac{C - 6\cos x}{(x + 3)^3}\)