Question

Solve 2 dy/dx+y tan⁡x = (4x+5)^2/cos⁡x y^3

Answer 1

Answer on Question #60706 - Math - Differential Equations

Question

Solve:

$$2 \frac {d y}{d x} + y \cdot \tan x = \frac {(4 x + 5) ^ {2}}{\cos x} y ^ {3}.$$

Solution

This is the Bernoulli equation:

$$2 y ^ {\prime} + y \tan x = \frac {(4 x + 5) ^ {2}}{\cos x} y ^ {3}. \quad (1)$$

We first notice that $y = 0$ is a solution. Division by $y^3$ yields:

$$\frac {2 y ^ {\prime}}{y ^ {3}} + \frac {\tan x}{y ^ {2}} = \frac {(4 x + 5) ^ {2}}{\cos x}. \quad (2)$$

Changing variables gives the equations:

$$z = \frac {1}{y ^ {2}}; \quad (3) \text{ and } z ^ {\prime} = - \frac {2 y ^ {\prime}}{y ^ {3}}; \quad (4)$$ $$- z ^ {\prime} + z \tan x = \frac {(4 x + 5) ^ {2}}{\cos x}. \quad (5)$$

We use the method of variation of parameters:

$$- z ^ {\prime} + z \tan x = 0; \quad (6)$$ $$\frac {d z}{z} = \tan x \, dx. \quad (7)$$

Because: $\tan x dx = \frac{\sin x \, dx}{\cos x} = -\frac{d(\cos x)}{\cos x}$, \quad (8)

$$\frac {d z}{z} = - \frac {d (\cos x)}{\cos x}; \quad (9)$$ $$\ln z = - \ln (\cos x) + \ln C; \quad (10)$$ $$z = \frac {C (x)}{\cos x}. \quad (11)$$

Differentiation yields: $z' = \frac{C'(x)}{\cos x} + \frac{C(x) \sin x}{\cos^2 x}. \quad (12)$

Substituting (11) and (12) back into equation (5):

$$- \frac {C ^ {\prime} (x)}{\cos x} - \frac {C (x) \sin x}{\cos^ {2} x} + \frac {C (x) \sin x}{\cos^ {2} x} = \frac {(4 x + 5) ^ {2}}{\cos x}; \quad (13)$$ $$- \frac {C ^ {\prime} (x)}{\cos x} = \frac {(4 x + 5) ^ {2}}{\cos x}. \quad (14)$$ $$C ^ {\prime} (x) = - (4 x + 5) ^ {2}. \quad (15)$$

Integrate (15) to find $C(x)$ :

$$C (x) = - \int (4 x + 5) ^ {2} d x = - \frac {1}{4} \int (4 x + 5) ^ {2} d (4 x) = - \frac {1}{4} \int (4 x + 5) ^ {2} d (4 x +$$ $$5) = - \frac {1}{4 \cdot 3} \int 3 (4 x + 5) ^ {2} d (4 x + 5) = - \frac {1}{12} (4 x + 5) ^ {3} + \tilde {C} = - \frac {1}{12} ((4 x + 5) ^ {3} + C),$$ $$C = \text{const}.$$ $$C (x) = - \frac {1}{12} ((4 x + 5) ^ {3} + C), C = \text{const}. \quad (16)$$

Substituting (16) back into equation (11):

$$z = - \frac {1}{12} \cdot \frac {(4 x + 5) ^ {3} + C}{\cos x}; \quad (17)$$

Substituting (3) back into equation (17):

$$\frac {1}{y ^ {2}} = - \frac {1}{12} \cdot \frac {(4 x + 5) ^ {3} + C}{\cos x}; \quad (18)$$ $$y ^ {2} = - \frac {12 \cos x}{(4 x + 5) ^ {3} + C} = \frac {12 \cos x}{C _ {1} - (4 x + 5) ^ {3}}; \quad (19)$$ $$y = \pm 2 \sqrt {\frac {3 \cos x}{C _ {1} - (4 x + 5) ^ {3}}}. \quad (20)$$

Answer:

$$y = \pm 2 \sqrt {\frac {3 \cos x}{C _ {1} - (4 x + 5) ^ {3}}}; y = 0.$$

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