Advanced Engineering Math. Problem set 1.4 : 1 ~ 3

 뒤늦게 정리하는 Problem set 1.4 : 1 ~ 3

종이에 스캔하지 말고 이번엔 수식을 직접 적어서 포스팅한다.

 

Test for exactness, If not, use an integrating factors.

 

Problem 1.

Find Solution : $2xydx+x^{2}dy=0$

 

Assume that : 

$$M=2xy,N=x^2$$

 

By the assumption of continuity the two second partial derivatives are equals.

 

$$\frac{\partial M}{\partial y}=2x$$

$$\frac{\partial N}{\partial x}=2x$$

 

Because of M = N, The equation is exact.

 

Let, $u=\int Ndy+l(x)$

 

$$u=\int 2xdy+l(x)$$

$$u=2xy+l(x)$$

 

Differentiate with respect to x : 

 

$$\frac{du}{dx}=2y+l^{\prime }(x)=2xy$$

$$l^{\prime }(x)=2xy-2y$$

$$l(x)=x^{2}y-2xy+c$$

$$\therefore u(x,y)=2xy+x^{2}y-2xy+c=0$$

$$\therefore x^{2}y=c$$

 

Problem 2.

$x^{3}dx+y^{3}dy=0$

Let, $M=x^3,N=y^3$

The equation is extact ($\because \frac{\partial M}{\partial y}=0,\frac{\partial N}{\partial x}=0$)

Integrate, 

 

$$u=\int Mdx+k(y)$$

$$u=\frac{1}{4}{x}^4+k(y)$$

 

Differentiate with respect to y : 

 

$$\frac{\partial u}{\partial y}=k^{\prime }(y)=N=y^3$$

$$\therefore k(y)=\frac{1}{4}{y}^4$$

$$\therefore \frac{1}{4}{x}^4+\frac{1}{4}{y}^4=c$$

 

Problem 3.

$sinxcosydx+cosxsinydy=0$

 

The equation is extact ($\because \frac{\partial M}{\partial y}=-sinxsiny,\frac{\partial N}{\partial x}=-sinxsiny$)

Integrate, 

 

$$u=\int Mdx+k(y)$$

$$u=\int sinxcosydx+k(y)$$

$$u=-cosxcosy+k(y)$$

 

Differentiate with respect to y : 

 

$$\frac{\partial u}{\partial y}=cosxsiny+k^{\prime }(y)=cosxsiny$$

$$\therefore k(y)=c$$

$$\therefore cosxcosy=c$$