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 $A^{\prime }\left(x\right)=$

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