Question

solve dy/dx=-1-x^2+y^2

Answer 1

Answer to Question #89430 – Math – Differential Equations

Question

solve dy/dx=-1-x^2+y^2

Solution

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

As the above equation is in the form of the Ricotta Equation, So Its particular solution is given as:

$$\begin{array}{l} \frac{dy}{dx} = A(x) y^2 + B(x) y + C(x) \\ Y = y_1 + \frac{1}{v} \\ \frac{dy}{dx} = \frac{dy_1}{dx} - \frac{1}{v^2} \frac{dv}{dx} \\ \frac{dy}{dx} = y^2 - (1 + x^2) \end{array}$$

Here,

$$A(x) = 1 \quad B(x) = 0 \quad C(x) = -(x^2 + 1)$$

$Y_1 = -x$ is the particular solution of the differential equation because it is verifying the differential equation

For checking $y = -x$

So $y' = -1$

$$\begin{array}{l} Y' = y^2 - (x^2 + 1) \\ -1 = (-x)^2 - (x^2 + 1) \\ -1 = -1 \quad \text{checked and verified} \\ Y = y' + \frac{1}{v} \\ y = -x + \frac{1}{v} \\ \frac{dy}{dx} = -1 - \frac{1}{v^2} \frac{dv}{dx} \\ \frac{dy}{dx} = y^2 - (x^2 + 1) \\ -1 - \frac{1}{v^2} \frac{dv}{dx} = \left(\frac{1}{v} - x^2\right) \cdot 2 - x^2 - 1 \\ -1 - \frac{1}{v^2} \frac{dv}{dx} = \frac{1}{v^2} - \frac{2x}{v} + x^2 - x^2 + 1 \\ \end{array}$$

After Simplifying we have,

$$\begin{array}{l} \frac{dv}{dx} = -1 + 2xv \\ \frac{dv}{dx} + v(-2x) = -1 \quad \text{... Equation (1)} \\ \text{It is the form of} \quad \frac{dv}{dx} + vP(x) = Q(x) \\ \end{array}$$

Integrating Factor $= e^{\int P(x)dx}$

$$\begin{array}{l} = e^{\int -2x\,dx} \\ = e^{-x^2} \\ \end{array}$$

Multiplying Integration factor by equation(1)

$$e^{-x^2} \left(\frac{dv}{dx} - 2xv\right) = -e^{-x^2}$$ $$\begin{array}{l} \frac{d}{dx}(v.e^{-x2}) = -e^{-x2}\left(\frac{d}{dx}(V.I.F) = f(x)\right) \\ \int \frac{d}{dx}(v.e^{-x2}) = \int -e^{-x2} dx \\ (v.e^{-x2}) = \frac{1}{2}\sqrt{\pi} \operatorname{Erf}(x) + c \\ V = \frac{1}{x+y} \\ \frac{1}{x+y}. e^{-x^2} = \frac{1}{2}\sqrt{\pi} \operatorname{Erf}(x) + c \\ \end{array}$$

So, we have

$$y = \frac{e^{-x2}}{c - \frac{1}{2}\sqrt{\pi} \operatorname{Erf}(x)} - x$$

Answer provided by https://www.AssignmentExpert.com