Question

The general solution of the first order linear differential equation of the type $\frac{dy}{dx}+Py=Q$ is

• A. $y=e^{−∫Qdx}â‹\dots ∫\left(Q{e}^{Qdx}\right)dx+C$
• B. $y=e^{∫Pdx}â‹\dots ∫\left(Q{e}^{∫Pdx}\right)dx+C$
• C. $y=e^{−∫Pdx}â‹\dots ∫\left(Q{e}^{∫Pdx}\right)dx+C$
• D. None of the above

Answer 1

Answer: (C)

To solve the first order linear differential equation of the type
$\frac{dy}{dx}+Py=Q....$(i)
On multiplying both sides of the equation by a function of $x$ say $g(x)$ to get
$g(x)\frac{dy}{dx}+Pâ‹\dots (g(x))y=Qâ‹\dots g(x)$...(ii)
Choose $g(x)$ in such a way that $RHS$ becomes a derivative of $yâ‹\dots g(x)$.
i.e., $g(x)\frac{dy}{dx}+Pâ‹\dots g(x)y=\frac{d}{dx}[yâ‹\dots g(x)]$
$⇒g(x)\frac{dy}{dx}+Pâ‹\dots g(x)y=g(x)\frac{dy}{dx}+y{g}^{′}(x)$
$⇒P(x)=g^{′}(x)$
or $P=\frac{g^{′}(x)}{g(x)}$
Integrating both sides w.r.t. $x$, we get
$∫Pdx=∫\frac{g^{′}(x)}{g(x)}dx$
$⇒∫Pâ‹\dots dx=\log (g(x))$
$⇒g(x)=e^{∫Pdx}$
On multiplying the Eq. (i) by $g(x)=e^{∫Pdx}$, the LHS becomes the derivative of some function of $x$ and $y$. This function $g(x)=e^{∫Pdx}$ is called Integrating Factor (IF) of the given differential equation.
On substituting the value of $g(x)$ in Eq. (ii), we get
$e^{∫Pdx}\frac{dy}{dx}+P{e}^{∫Pdx}y=Qâ‹\dots e^{∫Pdx}$
$⇒\frac{d}{dx}\left(y{e}^{∫Pdx}\right)=Q{e}^{∫Pdx}$
On integrating both sides w.r.t. $x$, we get
$yâ‹\dots e^{∫Pdx}=∫\left(Qâ‹\dots e^{∫Pdx}\right)dx$
$⇒y=e^{−∫Pdx}â‹\dots ∫\left(Qe{e}^{∫Pdx}\right)dx+C$
which is the general solution of the differential equation.