The general solution of a differential equation of the type $\frac{dx}{dy}+P_{1}x=Q_{1}$ is

Question

Accepted Answer

The given differential equation is of the form $\frac{dx}{dy} + P_1 x = Q_1$. This is a first-order linear differential equation in terms of $x$ and $y$. To solve such an equation, we use the method of integrating factors.

### Explanation for Option 3:

The standard form of a first-order linear differential equation is:

$$
\frac{dx}{dy} + P_1 x = Q_1
$$

To solve this, we multiply through by an integrating factor, which is $e^{\int P_1 \, dy}$. The purpose of the integrating factor is to make the left-hand side of the equation an exact derivative. The equation then becomes:

$$
e^{\int P_1 \, dy} \frac{dx}{dy} + e^{\int P_1 \, dy} P_1 x = e^{\int P_1 \, dy} Q_1
$$

This can be rewritten as:

$$
\frac{d}{dy} \left( x \cdot e^{\int P_1 \, dy} \right) = Q_1 \cdot e^{\int P_1 \, dy}
$$

Integrating both sides with respect to $y$, we get:

$$
x \cdot e^{\int P_1 \, dy} = \int \left( Q_1 \cdot e^{\int P_1 \, dy} \right) \, dy + c
$$

This matches exactly with Option 3:

$$ 
x \cdot e^{\int P_1 \, dy} = \int \left( Q_1 \cdot e^{\int P_1 \, dy} \right) \, dy + c 
$$

### Explanation for why other options are incorrect:

- **Option 1**: $ye^{\int P_1dy}=\int(Q_{1}e^{\int P_1dy})dy+c$

This option incorrectly uses $y$ instead of $x$ in the expression. The differential equation is solved for $x$ as a function of $y$, not $y$ as a function of $x$.

- **Option 2**: $y.e^{\int P_{1}dx}=\int(Q_{1}e^{\int P_{1}dx})dx+c$

This option incorrectly uses $dx$ instead of $dy$ in the integrals. The equation is given in terms of $\frac{dx}{dy}$, so the integration should be with respect to $y$, not $x$.

- **Option 4**: $x.e^{\int P_{1}dx}=\int(Q_{1}e^{\int P_{1}dx})dx+c$

Similar to Option 2, this option incorrectly uses $dx$ instead of $dy$ in the integrals. The integration should be performed with respect to $y$ as per the given differential equation.

In conclusion, Option 3 is the correct solution form for the given differential equation, as it correctly applies the method of integrating factors to solve for $x$ in terms of $y$.