Question:
(1) Find the general solution of
\(\frac{dy}{dx}+P(x)y=Q(x)\), where \(P(x)\) and \(Q(x)\) are functions of \(x\).
(2) Solve the following differential equation:
\(x\frac{dy}{dx}+2y=\sin x\)
Answer:
(1) Multiply \(e^{\int P(x)dx}\) by both sides.
\[ e^{\int P(x)dx}\frac{dy}{dx}+P(x)e^{\int P(x)dx}y=e^{\int P(x)dx}Q(x) \]
We can notice that the left hand side can be modified as follows:
\[ \frac{d}{dx}\left(e^{\int P(x)dx}y\right) = e^{\int P(x)dx}Q(x) \]
Integrate both sides in terms of \(x\).
\[ e^{\int P(x)dx}y = \int e^{\int P(x)dx}Q(x)dx + C \]
\[ y = e^{-\int P(x)dx}\left(\int e^{\int P(x)dx}Q(x)dx + C \right) \]
The other solution of (1) When \(Q(x) = 0\), the solution is \(y=Ce^{-\int P(x)dx}\). Therefore, when \(Q(x) \neq 0\), the constant \(C\) can be a function of \(x\) and then find the \(C(x)\). \(y=Ce^{-\int P(x)dx}\) can be plugged in the original equation:
\[ \frac{d}{dx}Ce^{-\int P(x)dx}+P(x)y = Q(x) \]
\[ C’e^{-\int P(x)dx}-CP(x)e^{-\int P(x)dx}+P(x)y = Q(x) \]
\[ C’e^{-\int P(x)dx}-P(x)y+P(x)y=Q(x) \]
\[ C’=Q(x)e^{\int P(x)dx} \]
\[ C(x) = \int Q(x)e^{\int P(x)dx}dx+C \]
Since \(C(x)=e^{\int P(x)dx}y\), we have
\[ e^{\int P(x)dx}y = \int Q(x)e^{\int P(x)dx}dx+C \]
\[ y = e^{-\int P(x)dx}\left(\int Q(x)e^{\int P(x)dx}dx+C \right) \]
(2) The given equation can be modified as
\[ \frac{dy}{dx}+\frac{2y}{x}=\frac{\sin x}{x} \]
From the previous discussion, we can let \(P(x)=2/x\) and \(Q(x)=\sin x/x\). So plug them into the result in (1).
\[ y = e^{-\int \frac{2}{x}dx}\left(\int \frac{\sin x}{x}e^{\int \frac{2}{x}dx}dx+C \right) \]
Think about the function \(f(x)=e^{-\int \frac{2}{x}dx}\). This becomes \(f(x)=e^{-2\ln x}\). In order to simplify it, take log of both sides.
\[ \ln f(x) = \ln e^{-2\ln x} \]
\[ \ln f(x) = -2\ln x \]
\[ \ln f(x) = \ln x^{-2} \]
\[ f(x) = x^{-2} \]
Thus, we have
\[ y = \frac{1}{x^2}\left(\int \frac{\sin x}{x}x^2 dx+C \right) \]
For the final steps,
\[ y = \frac{1}{x^2}\left(\int x\sin x dx+C \right) \]
\[ = \frac{1}{x^2}\left([-x\cos x + \int\cos x dx]+C \right) \]
\[ = \frac{1}{x^2}\left(-x\cos x + \sin x + C \right) \]