Answer to Question #296950 in Differential Equations for adarsh

Question

(a) The differential equation (2𝑥 2 + 𝑏𝑦 2 )𝑑𝑥 + 𝑐𝑥𝑦 𝑑𝑦 = 0 can made exact by multiplying with integrating factor 1 𝑥 ⁄ 2.

Then find the relation between 𝑏 and 𝑐. (b) Find one fourth roots of unity

Answer 1

(a)

$$(\dfrac{1}{x^2})(2x^2+by^2)dx+(\dfrac{1}{x^2})cxydy=0$$

$$(2+b\dfrac{y^2}{x^2})dx+c(\dfrac{y}{x})dy=0$$

$$\dfrac{\partial}{dx}(c(\dfrac{y}{x}))=\dfrac{\partial}{\partial y}(2+b\dfrac{y^2}{x^2})$$

$$-c(\dfrac{y}{x^2})=2b(\dfrac{y}{x^2})$$

$$c=-2b$$

(b) The polar form of $1$  is $\cos(0)+i\sin(0).$

$$k=0, \sqrt[4]{1}(\cos(\dfrac{0+2\pi (0)}{4})+i\sin(\dfrac{0+2\pi (0)}{4}))=1$$

$$k=1, \sqrt[4]{1}(\cos(\dfrac{0+2\pi (1)}{4})+i\sin(\dfrac{0+2\pi (1)}{4}))=i$$

$$k=2, \sqrt[4]{1}(\cos(\dfrac{0+2\pi (2)}{4})+i\sin(\dfrac{0+2\pi (2)}{4}))=-1$$

$$k=3, \sqrt[4]{1}(\cos(\dfrac{0+2\pi (3)}{4})+i\sin(\dfrac{0+2\pi (3)}{4}))=-i$$

$\sqrt[4]{1}=1$

$\sqrt[4]{1}=i$

$\sqrt[4]{1}=-1$

$\sqrt[4]{1}=-i$