Question

Write the ordinary differential equation ydx+(xy+x-3y)dy=0 in the linear form , and hence find its solution.

Answer 1

Answer on Question #66142 – Math – Differential Equations

Question

Write the ordinary differential equation $y dx + (xy + x - 3y) dy = 0$ in the linear form, and hence find its solution

Solution

We have the differential equation

$$y dx + (xy + x - 3y) dy = 0$$

Dividing equation (1) by $y dy$ we get

$$\frac{dx}{dy} + \left(1 + \frac{1}{y}\right)x = 3$$

This equation (2) has the linear form:

$$\frac{dx}{dy} + P(y)x = Q(y)$$

We need to multiply both sides by the integrating factor

$$I(y) = \exp\left(\int P(y) dy\right)$$

and integrate both sides. For this problem we have

$$I(y) = \exp\left(\int \left(1 + \frac{1}{y}\right) dy\right) = \exp(y + \ln y) = y e^{y}$$

Multiplying both sides of the differential equation (2) by $y e^{y}$, we get

$$y e^{y} \frac{dx}{dy} + y e^{y} \left(1 + \frac{1}{y}\right)x = 3 y e^{y}$$

or

$$y e^{y} \frac{dx}{dy} + e^{y}(y + 1)x = 3 y e^{y}$$

or

$$\frac{d}{dy} (x y e^{y}) = 3 y e^{y}$$

Integrating both sides, we get

$$x y e^{y} = 3 \int y e^{y} dy = 3 \left(y e^{y} - \int e^{y} dy\right) = 3 (y e^{y} - e^{y}) + C$$

Dividing equation by $y e^{y}$, we get a solution of the initial equation

$$x = 3 \left(1 - \frac{1}{y}\right) + \frac{C}{y} e^{-y}$$

Answer: $\frac{dx}{dy} + \left(1 + \frac{1}{y}\right)x = 3$; $x = 3 \left(1 - \frac{1}{y}\right) + \frac{C}{y} e^{-y}$.

Answer provided by https://www.AssignmentExpert.com