Question

If $y=A\left(x\right)e^{-\int Pdx} $ is a solution of $ \frac{dy}{dx}+P\left(x\right) y =Q\left(x\right) ,$ then $A'\left(x\right) = $

• A. $e^{\int Pdx}$
• B. $Q (x) e^{ - \int Pdx}$
• C. $\int Q (x) e^{ \int Pdx} dx $
• D. $Q (x) e^{ \int Pdx}$

Answer 1

Answer: (D)

Given, $y=A(x) e^{-\int P d x}$
Since, given differential equation is linear differential equation.
$\therefore IF =e^{\int P(x) d x}$
Now, solution of differential equation is
$y \cdot( IF )=\int Q(x) \times( IF ) d x $
$\Rightarrow y \times e^{\int P(x) d x}=\int Q(x) \cdot e^{\int F(x) d x} \cdot d x \dots$(i)
As mentioned above, $y e^{\int P(x) d x}=A(x)$
By Eq. (i), we get
$A(x)=\int Q(x) e^{\int P(x) d x} d x$
Then, $A'(x)=Q(x) e^{\int P(x) d x}$