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 : $ 2xy dx + x^2 dy = 0 $

 

Assume that : 

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

 

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

 

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

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

 

Because of M = N, The equation is exact.

 

Let, $ u = \int{N dy} + l(x) $

 

$$ u = \int{2x dy} + l(x) $$

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

 

Differentiate with respect to x : 

 

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

$$ l'(x) = 2xy - 2y $$

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

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

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

 

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, 

 

$$ u = \int{M dx} + k(y) $$

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

 

Differentiate with respect to y : 

 

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

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

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

 

Problem 3.

$ sin x cos y dx + cos x sin y dy = 0 $

 

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

Integrate, 

 

$$ u = \int{M dx} + k(y) $$

$$ u = \int sinx cos y dx + k(y) $$

$$ u = - cos x cos y + k(y) $$

 

Differentiate with respect to y : 

 

$$ \frac {\partial u} {\partial y} = cos x sin y + k'(y) = cos x sin y $$

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

$$ \therefore cos x cos y = c $$