Question

The general solution of the differential equation of the type \(\frac{dy}{dx} + py = Q\) is:
(where \(P\) and \(Q\) are functions of \(x\) only or constant)

• A. \(y\,e^{\int p\,dy} = \int \left(Q\,e^{\int p\,dy}\right) dy + c\) : (where \(c\) is an arbitrary constant)
• B. \(y\,e^{\int p\,dx} = \int \left(Q\,e^{\int p\,dx}\right) dx + c\) : (where \(c\) is an arbitrary constant)
• C. \(x\,e^{\int p\,dy} = \int \left(Q\,e^{\int p\,dy}\right) dy + c\) : (where \(c\) is an arbitrary constant)
• D. \(x\,e^{\int p\,dx} = \int \left(Q\,e^{\int p\,dx}\right) dx + c\) (where \(c\) is an arbitrary constant)

Hint:

Multiply by the integrating factor \(e^{\int P\,dx}\) and integrate with respect to \(x\).

Answer 1

The Correct Option is B

Step 1: Start from the product rule.
We want a function $u(x)$ such that $\frac{d}{dx}(u\,y) = u\left(\frac{dy}{dx} + Py\right)$. Expanding the left side, $u\,y' + u'\,y$, we need $u' = Pu$.

Step 2: Solve for u.
$\frac{u'}{u} = P$, so $\ln u = \int P\,dx$ and $u = e^{\int P\,dx}$.

Step 3: Use u in the equation.
Multiply the given equation by $u$: $\frac{d}{dx}(u\,y) = Q\,u$. Integrate with respect to $x$: $u\,y = \int Q\,u\,dx + c$.

Step 4: Match the option.
Putting $u = e^{\int P\,dx}$ back gives $y\,e^{\int P\,dx} = \int Q\,e^{\int P\,dx}\,dx + c$, which is option 2.

Final Answer:
Option 2 is correct. \[ \boxed{y\,e^{\int P\,dx} = \int Q\,e^{\int P\,dx}\,dx + c} \]