Question

2. Solve the initial value problem (1+y^2)dx+(1+x^2)d,y(0)=−1

Tan^−1 x−tan^−1 y=−π/4
Tan^−1 x+tan^−1 y=−π/4
Tan^−1 x+tan^−1 y=−π/4
-Tan^−1 x−tan^−1 y=−π/4

Answer 1

Answer on Question #79276 – Math – Differential Equations

Question

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

Solution

This equation with separating variables:

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

We integrate both sides of equation:

$$\int \frac{dx}{1 + x^{2}} = \int -\frac{dy}{1 + y^{2}}$$ $$\tan^{-1} x = -\tan^{-1} y + C$$ $$\tan^{-1} 0 = -\tan^{-1}(-1) + C$$ $$0 = -(-\frac{\pi}{4}) + C$$ $$C = -\frac{\pi}{4}$$

We substitute the obtained constant in equation:

$$\tan^{-1} x = -\tan^{-1} y - \frac{\pi}{4}$$ $$\tan^{-1} x + \tan^{-1} y = -\frac{\pi}{4}$$

Answer: $\tan^{-1} x + \tan^{-1} y = -\frac{\pi}{4}$.

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