Answer to Question #137931 in Differential Equations for Jan Ross

Question

1. y'-2y=y^2 ;y=3 when x=0
2. (y')^2=(1-y^2)/(1-x^2); y=1/2 when x=1
3. x(e^y)dy+((x^2)+1dx)/y=0

Answer 1

1. Rewrite the equation:

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

Integrate both parts of the equation:

$$\int \frac{dy}{(y+1)^2-1}= \int dx$$ $$\frac{1}{2} ln |\frac{y}{y+2}| =x+C$$ $$ln |\frac{y}{y+2}| =2x+C$$ $$\frac{y}{y+2}=e^{2x+C}$$ $$y=e^{2x+C}y+2e^{2x+C}$$ $$y(1-2e^{2x+C})=2e^{2x+C}$$ $$y=\frac{2e^{2x+C}}{1-2e^{2x+C}}$$

Substitute y=3, x=0:

$$3 = \frac{2e^C}{1-2e^C}$$ $$2e^C =3-6e^C$$ $$8e^C =3$$ $$e^C =\frac{3}{8}$$

Hence

$$y= \frac{2 \cdot \frac{3}{8}e^{2x}}{1-2 \cdot \frac{3}{8}e^{2x}}$$ $$y= \frac{ \frac{3}{4}e^{2x}}{1-\frac{3}{4}e^{2x}}$$ $$y= \frac{ 3e^{2x}}{4-3e^{2x}}$$

2. Rewrite the equation:

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

Integrate both parts of the equation:

$$\int \frac{dy}{\sqrt{1-y^2}}=\int \frac{dx}{\sqrt{1-x^2}}$$ $$arcsin y = arcsin x +C$$

Since $y = \frac{1}{2}, x = 1,$ then

$$arcsin \frac{1}{2} = arcsin {1} +C$$ $$C = \frac{\pi}{6}-\frac{\pi}{2} =-\frac{\pi}{3}$$

Hence

$$arcsin y = arcsin x -\frac{\pi}{3}$$ $$y = sin ( arcsin x -\frac{\pi}{3})$$ $$y = x cos \frac{\pi}{3}-sin \frac{\pi}{3} cos (arcsin x)$$ $$y = \frac{x}{2}-\frac{\sqrt {3}}{2}\cdot \sqrt {1-x^2}$$ $$y = \frac{1}{2}(x- \sqrt {3-3x^2})$$

3. Rewrite the equation:

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

Integrate both parts of the equation:

$$\int{y e^y dy} = -\int (x+\frac{1}{x})dx$$ $$y e^y -e^y = -\frac{x^2}{2}+ln x+C$$ $$e^y(y-1)= -\frac{x^2}{2}+ln x+C$$