Advanced Engineering Math. Problem set 2.7 : 7 ~ 9

Problem 7 ~ 9는 미분 연산자로 표현이 되어있다. 평소대로 읽어서 풀면 된다. 이런 표현은 자주 나오는데, 특히 공기역학 공부시 실체적 도함수(Substantial Derivatives)는 유용하니 익숙해지면 편하다.

 

Problem 7.

Find the solution

$$ y'' + 2 y' + \frac {3}{4} y = 3 e^{x} + \frac {9}{2} x $$

1) Solution of homogeneous ODE

 Let's solve the equation when right hand side is 0

By the characteristic equation, there are real roots, $ \lambda = - \frac {1}{2} or - \frac {3}{2} $

Hence, the solution $ y_{h} $ : 

$$ \therefore y_{h} = c_{1} e^{- \frac {1}{2} x} + c_{2} e^{- \frac{3}{2} x} $$

2) Solution of Nonhomogeneous ODE

 Let's assume the $ y_{p} $ is the $ C e^{x} + K_{1} x + K_{0} $ 

It is determined by Basic Rule and Sum Rule of method of undetermined coefficient.

$$ y_{p} = Ce^{x} + K_{1} x + K_{0} $$

$$ y'_{p} = Ce^{x} + K_{1} $$

$$ y''_{p} = Ce^{x} $$

Substitute into equation, 

$$ (1 + 2 + \frac{3}{4}) C e^{x} + \frac {3}{4} K_{1} x + 2K_{1} + \frac {3}{4} K_0 = 3e^{x} + \frac {9}{2} x $$

In short, 

$$ C = \frac{4}{5} \ , \ K_{1} = 6 \ , \ K_{0} = - 16 $$

Hence, the $ y_{p} $ is,

$$ \therefore y_{p} = \frac {4}{5} e^{x} + 6 x - 16 $$

The General Solution is,

$$ \therefore y(x) = C_{1} e^{- \frac{1}{2} x } + C_{2} e^{- \frac {3}{2} x} + \frac {4}{5} e^{x} + 6 x - 16 $$

 

 

Problem 8.

Find the solution

$$ 3 y'' + 27 y = 3 cos x + cos {3x} $$

Divide by 3, 

$$ y'' + 9 y = cos x + \frac{1}{3} cos {3x} $$

1) Solution of homogeneous ODE

 Let's solve the equation when the RHS is 0

By the characteristic equation, there are two complex roots, $ \lambda = \pm 3 i $

Therefore, the solution $ y_{h} $ : 

$$ \therefore y_{h} = C_{1} cos 3x + C_{2} sin 3x $$

2) Solution of Nonhomogeneous ODE

 Let's assume the $ y_{p} = A cos x + B sin x + C x cos {3x} + D x sin {3x} $

The terms " $ A cos x + B sin x $ " is determined by Basic Rule,

And another term is defined by Basic Rule and Modification Rule because of correspondence of $ y_{h} $.

Substitute into the equation, and we get,

$$ 8A cos x + 8 B sinx - 6C sin {3x} + 6D cos {3x} = cos x + \frac{1}{3} cos{3x} $$

$$ A = \frac {1}{8} \ , \ B = 0 \ , \ C = 0 \ , \ D = \frac {1}{18} $$

Hence, the $ y_{p} $ is,

$$ \therefore y_{p} = \frac{1}{8} cosx + \frac{1}{18} x sin 3x $$

The General Solution is, 

$$ \therefore y(x) = C_{1} cos {3x} + C_{2} sin {3x} + \frac {1}{8} cos x + \frac {1}{18} x sin {3x} $$

 

Problem 9.

Find the solution

$$ y'' - 16y = 9.6 e^{4x} + 30 e^{x} $$

1) Solution of homogeneous ODE

By the characteristic equation, $ \lambda = \pm 4 $

$$ \therefore y_{h} = c_{1} e^{4x} + c_{2} e^{-4x} $$

2) Solution of Nonhomogeneous ODE

We will use the Sum Rule.

when checking the first term of right hand side, we can apply the modification rule by mulfiplying x,

and second term of RHS will be deteremind Basic Rule.

$$ y_{p} = Axe^{4x} + Be^{x} $$

$$ y'_{p} = Ae^{4x} + 4 A x e^{4x} + B e^{x} $$

$$ y''_{p} = 4A e^{4x} + 4A e^{4x} + 16 Axe^{4x} + Be^{x} $$

Substitute into the equation, we get,

$$ 8A e^{4x} - 15B e^{x} = 9.6 e^{4x} + 30e^{x} $$

$$ A = 1.2 \ , \ B = -2 $$

Hence, the $ y_{p} $ is,

$$ \therefore y_{p} = 1.2 x e^{4x} - 2 e^{x} $$

The General Solution is,

$$ \therefore y(x) = c_{1} e^{4x} + c_{2} e^{-4x} + 1.2 x e^{4x} - 2 e^{x} $$