3. Solution (x^2y')' + (2xy)' = 0 x^2y' + 2xy = c_1 y' = dy/dx x^2dy + ydx^2 = c_1dx dyx^2 = c_1dx

(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 $x_1 = \frac{|A_1|}{|A|} = \frac{-3}{-3} = 1, x_2 = \frac{|A_2|}{|A|} = \frac{6}{-3} = -2, x_3 = \frac{|A_3|}{|A|} = \frac{-9}{-3} = 3$

3. Solution

\( (x^2y')' + (2xy)' = 0 \)

\( x^2y' + 2xy = c_1 \)

\( y' = dy/dx \)

\( x^2dy + ydx^2 = c_1dx \)

\( dyx^2 = c_1dx \)

相關申論題

相關試卷

3. Solution \( (x^2y')' + (2xy)' = 0 \) \( x^2y' + 2xy = c_1 \) \( y' = dy/dx \) \( x^2dy + ydx^2 = c_1dx \) \( dyx^2 = c_1dx \)

相關申論題

相關試卷

3. Solution

\( (x^2y')' + (2xy)' = 0 \)

\( x^2y' + 2xy = c_1 \)

\( y' = dy/dx \)

\( x^2dy + ydx^2 = c_1dx \)

\( dyx^2 = c_1dx \)