Question

Solve the differencial equation y" = 1+(y')2.

Answer 1

Answer on Question#40032 – Math – Differential Calculus | Equation

Solve the differential equation $y'' = 1 + (y')^2$.

Solution.

Let $v(x) = y'(x) \to y'' = v'$ and from our equation:

$$v' = v^2 + 1 \to \frac{v'}{v^2 + 1} = 1 \to \int \frac{v'}{v^2 + 1} \, dx = \int 1 \, dx \to \int \frac{1}{v^2 + 1} \, dv = \int 1 \, dx \to$$ $$\to \arctan(v) = x + c_1 \to v = \tan(x + c_1).$$

Then: $y' = v = \tan(x + c_1) \to y = \int \tan(x + c_1) \, dx \to y = -\ln(\cos(x + c_1) + c_2)$.

Answer: $y(x) = -\ln(\cos(x + c_1) + c_2$, where $c_1$ and $c_2$ are the arbitrary constants.