Question

solve the differencial equation:
y sin 2x dx= (1+y2+(cos)2 x)dy.

Answer 1

Answer on Question#40384 – Math – Differential Calculus

Solve the differential equation:

$$y \sin (2x) \, dx = (1 + y^2 + \cos(2x)) \, dy.$$

Solution:

We have

$$y \sin(2x) \, dx - (1 + y^2 + \cos(2x)) \, dy = 0,$$ $$M(x, y) \, dx + N(x, y) \, dy = 0,$$

where

$$M(x, y) = y \sin(2x), \quad N(x, y) = - (1 + y^2 + \cos(2x)).$$

Thus

$$\frac{\partial M}{\partial y} = \sin(2x), \quad \frac{\partial N}{\partial x} = 2 \sin(2x).$$

Because $\frac{\partial M}{\partial y} \neq \frac{\partial N}{\partial x}$ then we can try to find an integrating factor $m(x, y)$.

We get

$$\frac{\frac{\partial M}{\partial y} - \frac{\partial N}{\partial x}}{-M(x, y)} = \frac{\sin(2x) - 2 \sin(2x)}{-y \sin(2x)} = \frac{-\sin(2x)}{-y \sin(2x)} = \frac{1}{y}.$$

Because

$$\frac{\frac{\partial M}{\partial y} - \frac{\partial N}{\partial x}}{-M(x, y)}$$

is the function of variable $y$ only then we have next differential equation

$$\frac{1}{m(y)} \frac{dm}{dy} = \frac{\frac{\partial M}{\partial y} - \frac{\partial N}{\partial x}}{-M(x, y)},$$ $$\frac{dm}{m(y)} = \frac{dy}{y},$$ $$\int \frac{dm}{m(y)} = \int \frac{dy}{y},$$ $$\ln m(y) = \ln y,$$ $$\boxed{m(y) = y}$$

Thus we get next differential equation

$$(y \sin (2 x) d x - (1 + y ^ {2} + \cos (2 x)) d y) \cdot m (y) = 0,$$ $$y ^ {2} \sin (2 x) d x - (y + y ^ {3} + y \cos (2 x)) d y = 0,$$ $$M _ {1} (x, y) d x + N _ {1} (x, y) d y = 0,$$

where

$$M _ {1} (x, y) = y ^ {2} \sin (2 x), \qquad N _ {1} (x, y) = - (y + y ^ {3} + y \cos (2 x)).$$

Thus

$$\frac {\partial M _ {1}}{\partial y} = 2 y \sin (2 x), \frac {\partial N _ {1}}{\partial x} = 2 y \sin (2 x).$$

Because $\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}$ then we get differential equation in exact differentials. Thus we have

$$u (x, y) = \int M _ {1} (x, y) d x = \int y ^ {2} \sin (2 x) d x = - \frac {y ^ {2}}{2} \cos (2 x) + \varphi (y).$$

To find function $\varphi (y)$ we get next equation

$$\frac {\partial}{\partial y} \left(- \frac {y ^ {2}}{2} \cos (2 x) + \varphi (y)\right) = - (y + y ^ {3} + y \cos (2 x)),$$ $$- y \cos (2 x) + \frac {d \varphi}{d y} = - y - y ^ {3} - y \cos (2 x),$$ $$\frac {d \varphi}{d y} = - y - y ^ {3},$$ $$\varphi (y) = \int (- y - y ^ {3}) d y = - \frac {y ^ {2}}{2} - \frac {y ^ {4}}{4} + c.$$

So

$$u (x, y) = - \frac {y ^ {2}}{2} \cos (2 x) - \frac {y ^ {2}}{2} - \frac {y ^ {4}}{4} + c.$$

Thus the general solution of the differential equation is

$$- \frac {y ^ {2}}{2} \cos (2 x) - \frac {y ^ {2}}{2} - \frac {y ^ {4}}{4} + c = 0,$$ $$- \frac {y ^ {2}}{2} \cos (2 x) - \frac {y ^ {2}}{2} - \frac {y ^ {4}}{4} = - c,$$ $$\frac {y ^ {2}}{2} (1 + \cos (2 x)) + \frac {y ^ {4}}{4} = c,$$ $$2 y ^ {2} (1 + \cos (2 x)) + y ^ {4} = C$$

where $C = 4c$

Answer:

$$2 y ^ {2} (1 + \cos (2 x)) + y ^ {4} = C$$