Question

Find the solution of the Riccati equation:
dy/dx = 2cos^2 x - sin^2 x + y^2/ 2cosx ; y1(x) = sinx.

Answer 1

Answer on Question# #50671 – Math– Differential Calculus | Equations

Question

Find the solution of the Riccati equation: $\frac{dy}{dx} = 2\cos^2 x - \sin^2 x + \frac{y^2}{2\cos x}$ ; $y_1(x) = \sin x$ .

Note that there is an error in the form of given equation! The function $y_{1}(x) = \sin (x)$ is not a particular solution of equation

$$\frac {d y}{d x} = 2 \cos^ {2} (x) - \sin^ {2} (x) + \frac {y ^ {2}}{2 \cos (x)},$$ $$\left(\frac {d y _ {1}}{d x} \neq 2 \cos^ {2} (x) - \sin^ {2} (x) + \frac {y _ {1} ^ {2}}{2 \cos (x)}\right)$$

but it satisfies the equation

$$\frac {d y}{d x} = \frac {2 \cos^ {2} (x) - \sin^ {2} (x)}{2 \cos (x)} + \frac {1}{2 \cos (x)} y ^ {2},$$ $$\left(\frac {d y _ {1}}{d x} = \frac {2 \cos^ {2} (x) - \sin^ {2} (x)}{2 \cos (x)} + \frac {1}{2 \cos (x)} y _ {1} ^ {2}\right).$$

So we will solve the equation $(^{*})$

Solution

The Riccati equation is the first-order nonlinear ordinary differential equation. In the general case it is an equation of the type

$$\frac {d y}{d x} = P (x) y ^ {2} + Q (x) y + R (x)$$

where $P(x), Q(x), R(x)$ are continuous functions. In our case

$$P (x) = \frac {1}{2 \cos (x)}, \qquad Q (x) = 0, \qquad R (x) = \frac {2 \cos^ {2} (x) - \sin^ {2} (x)}{2 \cos (x)}$$

and equation (1) takes the form

$$\frac {d y}{d x} = \frac {1}{2 \cos (x)} y ^ {2} + \frac {2 \cos^ {2} (x) - \sin^ {2} (x)}{2 \cos (x)}.$$

If we know a particular solution of the Riccati equation, then its general solution is

$$y (x) = y _ {1} (x) + u (x),$$

where $u(x)$ is a new unknown function. In our case the particular solution is $y_{1}(x) = \sin (x)$ .

Let us solve equation (2) knowing the particular solution $y_{1}(x)$ .

Substituting (3) into (2) we have

$$\frac {d}{d x} (y _ {1} + u) = \frac {1}{2 \cos (x)} (y _ {1} + u) ^ {2} + \frac {2 \cos^ {2} (x) - \sin^ {2} (x)}{2 \cos (x)}$$ $$\frac {d y _ {1}}{d x} + \frac {d u}{d x} = \frac {1}{2 \cos (x)} \left(y _ {1}\right) ^ {2} + \frac {2 y _ {1} u}{2 \cos (x)} + \frac {u ^ {2}}{2 \cos (x)} + \frac {2 \cos^ {2} (x) - \sin^ {2} (x)}{2 \cos (x)}$$

As $y_{1}(x)$ is a function that satisfies the equation (2), then

$$\frac {d y _ {1}}{d x} = \frac {1}{2 \cos (x)} y _ {1} ^ {2} + \frac {2 \cos^ {2} (x) - \sin^ {2} (x)}{2 \cos (x)}$$

Thus, from (4), taking into account (5), we obtain the differential equation for the function $u(x)$

$$\frac {d u}{d x} = \frac {2 y _ {1} u}{2 \cos (x)} + \frac {u ^ {2}}{2 \cos (x)}$$

It is the Bernoulli equation. The substitution

$$z (x) = \frac {1}{u (x)}.$$

transforms the equation (6) into a linear differential equation admitting integration:

$$\frac {d z}{d x} + z \tan (x) = - \frac {1}{2 \cos (x)}.$$

Further, to find the solution of equation (8) we use the variation of constants method:

$$\frac {d z}{d x} + z \tan (x) = 0 \Rightarrow$$ $$\frac {d z}{z} + \frac {\sin (x)}{\cos (x)} d x = 0 \Rightarrow$$ $$\ln | z | - \ln | \cos (x) | = \ln | C | \Rightarrow$$ $$z (x) = C (x) \cos (x).$$ $$\frac {d}{d x} (C (x) \cos (x)) + (C (x) \cos (x)) \tan (x) = - \frac {1}{2 \cos (x)} \Rightarrow$$ $$\left(C ^ {\prime} (x) \cos (x) - C (x) \sin (x)\right) + \left(C (x) \cos (x)\right) \frac {\sin (x)}{\cos (x)} = - \frac {1}{2 \cos (x)} \Rightarrow$$ $$C ^ {\prime} (x) \cos (x) - C (x) \sin (x) + C (x) \sin (x) = - \frac {1}{2 \cos (x)} \Rightarrow$$ $$C ^ {\prime} (\mathrm {x}) \cos (\mathrm {x}) = - \frac {1}{2 \cos (x)} \Rightarrow$$ $$C ^ {\prime} (\mathrm {x}) = - \frac {1}{2 (\cos (x)) ^ {2}} \Rightarrow$$ $$C (x) = - \int {\frac {d x}{2 (\cos (x)) ^ {2}} = - \frac {\tan (x)}{2} + C _ {1}} \Rightarrow$$ $$z(x) = C(x) \cos(x) = \left(- \frac {\tan(x)}{2} + C _ {1}\right) \cos(x) = C _ {1} \cos(x) - \frac {1}{2} \sin(x).$$ $$z(x) = C_{1} \cos(x) - \frac{1}{2} \sin(x) = \frac{1}{2} \left(\overline{C_{1}} \cos(x) - \sin(x)\right), \quad \overline{C}_{1} = 2 C_{1},$$ $$z(x) = \frac{1}{2} \left(\overline{C_{1}} \cos(x) - \sin(x)\right).$$

Hence, using (7) and (3) we get solution of equation (2):

$$y(x) = \sin(x) + \frac {2}{\overline{C_{1}} \cos(x) - \sin(x)},$$

where $\overline{C_1}$ is an arbitrary real constant.

Let us check whether the function $y(x)$ satisfies the equation (2):

$$\frac {d}{d x} \Big (\sin(x) + \frac {2}{\overline{C_{1}} \cos(x) - \sin(x)} \Big) = \frac {1}{2 \cos(x)} \Big (\sin(x) + \frac {2}{\overline{C_{1}} \cos(x) - \sin(x)} \Big) ^ {2} + \frac {2 \cos^{2}(x) - \sin^{2}(x)}{2 \cos(x)},$$ $$\left(\cos(x) + \frac {2 (\overline{C_{1}} \sin(x) + \cos(x))}{(\overline{C_{1}} \cos(x) - \sin(x))^{2}}\right) = \frac {1}{2 \cos(x)} \Big (\sin^{2}(x) + \frac {4 \sin(x)}{\overline{C_{1}} \cos(x) - \sin(x)} + \frac {4}{(\overline{C_{1}} \cos(x) - \sin(x))^{2}} \Big) + \frac {2 \cos^{2}(x) - \sin^{2}(x)}{2 \cos(x)},$$ $$\left(\cos(x) + \frac {2 (\overline{C_{1}} \sin(x) + \cos(x))}{(\overline{C_{1}} \cos(x) - \sin(x))^{2}}\right) = \frac {\sin^{2}(x)}{2 \cos(x)} + \frac {4 \sin(x)}{2 \cos(x) (\overline{C_{1}} \cos(x) - \sin(x))} + \frac {4}{2 \cos(x) (\overline{C_{1}} \cos(x) - \sin(x))^{2}} +$$ $$\frac {2 \cos^{2}(x) - \sin^{2}(x)}{2 \cos(x)},$$ $$\frac {2 (\overline{C_{1}} \sin(x) + \cos(x))}{(\overline{C_{1}} \cos(x) - \sin(x))^{2}} = \frac {4 \sin(x)}{2 \cos(x) (\overline{C_{1}} \cos(x) - \sin(x))} + \frac {4}{2 \cos(x) (\overline{C_{1}} \cos(x) - \sin(x))^{2}},$$ $$\frac {2 (\overline{C_{1}} \sin(x) + \cos(x))}{(\overline{C_{1}} \cos(x) - \sin(x))^{2}} = \frac {2}{\cos(x)} \left(\frac {\sin(x) (\overline{C_{1}} \cos(x) - \sin(x)) + 1}{(\overline{C_{1}} \cos(x) - \sin(x))^{2}}\right),$$ $$\frac {2 (\overline{C_{1}} \sin(x) + \cos(x))}{(\overline{C_{1}} \cos(x) - \sin(x))^{2}} = \frac {2}{\cos(x)} \left(\frac {\overline{C_{1}} \sin(x) \cos(x) + \cos^{2}(x)}{(\overline{C_{1}} \cos(x) - \sin(x))^{2}}\right),$$ $$\frac {2 (\overline{C_{1}} \sin(x) + \cos(x))}{(\overline{C_{1}} \cos(x) - \sin(x))^{2}} = \frac {2 (\overline{C_{1}} \sin(x) + \cos(x))}{(\overline{C_{1}} \cos(x) - \sin(x))^{2}}.$$

As we see, the function (10) satisfies the equation (2).

Answer: $y(x) = \sin(x) + \frac{2}{\overline{C_1} \cos(x) - \sin(x)}$.

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