Question

The general solution of a differential equation of the type $\frac{dx}{dy}+P_{1}x=Q_1$ is
(a) $y{e}^{\int P_{1}dy}=\int \left\{Q_1{e}^{\int P_{1}dy}\right\}dy+C$

(b) $y{e}^{\int P_{1}dx}=\int \left\{Q_1{e}^{\int P_{1}dx}\right\}dx+C$

(c) $x{e}^{\int P_{1}dy}=\int \left\{Q_1{e}^{\int P_{1}dy}\right\}dy+C$

(d) $x{e}^{\int P_{1}dx}=\int \left\{Q_1{e}^{\int P_{1}dx}\right\}dx+C$

Answer 1

$\left(\mathrm{c}\right)x{e}^{\int P_{1}dy}=\int \left\{Q_1{e}^{\int P_{1}dy}\right\}dy+C$


We have,
$\frac{dx}{dy}+P_{1}x=Q_1$
Comparing with the equation $\frac{dx}{dy}+Px=Q$, we get
P = P₁
Q = Q₁
The general solution of the equation $\frac{dx}{dy}+Px=Q$ is given by
$x{e}^{\int Pdy}=\int \left\{Q{e}^{\int Pdy}\right\}dy+C$ ...(1)
Putting the value of P and Q in (1), we get
$x{e}^{\int P_{1}dy}=\int \left\{Q_1{e}^{\int P_{1}dy}\right\}dy+C$