Question

Solve the following ordinary differential equations:
i)

sin [ ] ( ) ln )(cos ) 0
1
ydx + x y + y dy =
x

Answer 1

Answer on Question #62061 – Math – Differential Equations

Question

Solve the following ordinary differential equations:

i)

$$\frac {1}{x} \sin y \, dx + [ (\ln x) (\cos y) + y ] \, dy = 0$$

Solution

Let us split the left-hand side of the equation in two parts

$$\left(\frac {\sin y}{x} \, dx + (\ln x) (\cos y) \, dy\right) + y \, dy = 0.$$

Consider

$$\frac {\sin y}{x} \, dx = \sin y \, d(\ln x); \, (\ln x) (\cos y) \, dy = \ln x \, d(\sin y);$$ $$y \, dy = d \frac {y ^ {2}}{2}.$$

Then

$$\frac {\sin y}{x} \, dx + (\ln x) (\cos y) \, dy = \sin y \, d(\ln x) + \ln x \, d(\sin y) = d (\sin y \cdot \ln x)$$

and the equation (1) is equivalent to

$$d (\sin y \cdot \ln x) + y \, dy = 0.$$

Integrating the formula (2) we get

$$\sin y \cdot \ln x + \frac {y ^ {2}}{2} = c,$$

where $c$ is an arbitrary real constant.

**Answer**: $\sin y\cdot \ln x + \frac{y^2}{2} = c.$

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