4. Solution dy + x y dx = 2 x x dx Integrate factor I(x) = e^∫ x dx = e^- x = ^-1 x d(^-1 x y) = 2 x dx ^-1 x y = -2 x + c_1

(a) The matrix of coefficients A and column matrix of constants B are $A = \begin{bmatrix} 1 & 3 & 1 \\ 2 & 5 & 1 \\ 1 & 2 & 3 \end{bmatrix}$ and $B = \begin{bmatrix} -2 \\ -5 \\ 6 \end{bmatrix}$ It is found that |A| = –3 ≠ 0. Thus Cramer’s rule can be applied. We get $A_1 = \begin{bmatrix} -2 & 3 & 1 \\ -5 & 5 & 1 \\ 6 & 2 & 3 \end{bmatrix}$ $A_2 = \begin{bmatrix} 1 & -2 & 1 \\ 2 & -5 & 1 \\ 1 & 6 & 3 \end{bmatrix}$ $A_3 = \begin{bmatrix} 1 & 3 & -2 \\ 2 & 5 & -5 \\ 1 & 2 & 6 \end{bmatrix}$ giving $|A_1| = -3, |A_2| = 6, |A_3| = -9$ 。Cramer’s rule now gives $x_1 = \frac{|A_1|}{|A|} = \frac{-3}{-3} = 1, x_2 = \frac{|A_2|}{|A|} = \frac{6}{-3} = -2, x_3 = \frac{|A_3|}{|A|} = \frac{-9}{-3} = 3$

4. Solution

\( dy + \tan x y dx = 2 \sin x \cos x dx \)

Integrate factor \( I(x) = e^{\int \tan x dx} = e^{-\ln \cos x} = \cos^{-1} x \)

\( d(\cos^{-1} x y) = 2 \sin x dx \)

\( \cos^{-1} x y = -2 \cos x + c_1 \)

相關申論題

相關試卷

4. Solution \( dy + \tan x y dx = 2 \sin x \cos x dx \) Integrate factor \( I(x) = e^{\int \tan x dx} = e^{-\ln \cos x} = \cos^{-1} x \) \( d(\cos^{-1} x y) = 2 \sin x dx \) \( \cos^{-1} x y = -2 \cos x + c_1 \)

相關申論題

相關試卷

4. Solution

\( dy + \tan x y dx = 2 \sin x \cos x dx \)

Integrate factor \( I(x) = e^{\int \tan x dx} = e^{-\ln \cos x} = \cos^{-1} x \)

\( d(\cos^{-1} x y) = 2 \sin x dx \)

\( \cos^{-1} x y = -2 \cos x + c_1 \)