Question

Solve the initial value problem (1+y2)dx+(1+x2)d,y(0)=−1

Answer 1

Answer on Question #69629 – Math – Differential Equations

Question

Solve the initial value problem $(1 + y^{2})dx + (1 + x^{2})dy = 0, y(0) = -1$.

Solution

$$(1 + y^{2})dx + (1 + x^{2})dy = 0,$$ $$y(0) = -1.$$

The equation (1) is separable because dividing (1) by $(1 + y^{2})(1 + x^{2})$ the variables can be separated:

$$\frac{dx}{1 + x^{2}} + \frac{dy}{1 + y^{2}} = 0,$$ $$\frac{dy}{1 + y^{2}} = -\frac{dx}{1 + x^{2}}.$$

Integrating both sides we get

$$\int \frac{dy}{1 + y^{2}} = \int -\frac{dx}{1 + x^{2}},$$ $$\tan^{-1} y = -\tan^{-1} x + C.$$

When $x = 0$, $y(0) = -1$:

$$\tan^{-1}(-1) = -\tan^{-1}(0) + C,$$ $$-\frac{\pi}{4} = 0 + C,$$ $$C = -\frac{\pi}{4},$$ $$\tan^{-1} y = -\tan^{-1} x - \frac{\pi}{4}.$$

Take the tangent of both sides of equality (2):

$$\tan(\tan^{-1} y) = \tan\left(-\tan^{-1} x - \frac{\pi}{4}\right),$$ $$y = -\tan\left(\tan^{-1} x + \frac{\pi}{4}\right),$$ $$y = -\frac{\tan(\tan^{-1} x) + \tan\left(\frac{\pi}{4}\right)}{1 - \tan(\tan^{-1} x) \cdot \tan\left(\frac{\pi}{4}\right)},$$ $$y = -\frac{x + 1}{1 - x},$$ $$y = \frac{x + 1}{x - 1}.$$

Answer: $y = \frac{x + 1}{x - 1}$.

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