MCQ
The general solution of differention eqution of the type $\frac{\text{dx}}{\text{dy}}+\text{P}_{1}\text{x}=\text{Q}_{1}$ is:
  • A
    $\text{ye}^{\int\text{P}_{1}\text{dy}}=\int \left\{\text{Q}_{1}\text{e}^{\int\text{P}_{1}\text{dy}}\right\}\text{dy}+\text{C}$
  • B
    $\text{ye}^{\int\text{P}_{1}\text{dy}}=\int \left\{\text{Q}_{1}\text{e}^{\int\text{P}_{1}\text{dx}}\right\}\text{dx}+\text{C}$
  • $\text{xe}^{\int\text{P}_{1}\text{dy}}=\int \left\{\text{Q}_{1}\text{e}^{\int\text{P}_{1}\text{dx}}\right\}\text{dx}+\text{C}$
  • D
    $\text{xe}^{\int\text{P}_{1}\text{dx}}=\int \left\{\text{Q}_{1}\text{e}^{\int\text{P}_{1}\text{dx}}\right\}\text{dx}+\text{C}$

Answer

Correct option: C.
$\text{xe}^{\int\text{P}_{1}\text{dy}}=\int \left\{\text{Q}_{1}\text{e}^{\int\text{P}_{1}\text{dx}}\right\}\text{dx}+\text{C}$
We have,
$\frac{\text{dx}}{\text{dy}}+\text{P}_{1}\text{x}=\text{Q}_{1}$
Comparing with the equation $\frac{\text{dx}}{\text{dy}}+\text{P}\text{x}=\text{Q}$ we get,
$\text{P}=\text{P}_{1}, \text{Q}=\text{Q}_{1}$
The solution of the equation $\frac{\text{dx}}{\text{dy}}+\text{P}\text{x}=\text{Q}$ is given by
$\text{xe}^{\int\text{P}_{1}\text{dy}}=\int \left\{\text{Q}_{1}\text{e}^{\int\text{P}_{1}\text{dy}}\right\}\text{dy}+\text{C}\ ...(\text{i})$
Putting the value of $P$ and $Q$ in $(i),$
$\text{xe}^{\int\text{P}_{1}\text{dy}}=\int \left\{\text{Q}_{1}\text{e}^{\int\text{P}_{1}\text{dy}}\right\}\text{dy}+\text{C}$

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