For the regular Sturm-Liouville problem "+= 0, 0 < x 0. Find upper and lower bounds for the leading eigenvalue of the regular Sturm-Liouville problem "x²+= 0, 0 < x <1, 26(0) - '(0) = 0, '(1) = 0.

x'(t) = of the any - 2 5-6 4 1-4-6|x (t) x (0) = -2 + -3 1 The only eigen value of this matrix is -3, a triple root. You must explicitly find matrix involved, with the exception cannot involve Also, your answer 0 imaginary any matrix inverses. number i.

Determine a Cauchy-Euler equation of minimum possible order with real coefficients if it 3 and m = -2i are two of the roots of its auxiliary equation. is known that m = You may find it helpful that a third-order Cauchy-Euler equation of the form ax y" + bx² y" + cxy + dy = 0 has the following auxiliary equation: am + (b – 3a)m² + (2a – b + c)m + d = 0.

Find a particular solution of ;) - (: :) (;) •(*)- -5 sint 4 0 -4t A Hint to minimize your computation: Eigenvalues of A are +2, C ) and A? = 41. A t+ cost Please type in a =. Then you obtain Yp 1+a sint

Determine a Cauchy-Euler equation of minimum possible order with real coefficients if it is known that m = 4 and m = -2i are two of the roots of its auxiliary equation. You may find it helpful that a third-order Cauchy-Euler equation of the form ax³y" + bx²y" + cxy + dy = 0 has the following auxiliary equation: am³ (b3a)m² + (2a-b+ c)m + d = 0.

find the eigenvalues 3*3 matrix (-1 4 -2 -3 4  0 -3  1 3)

Consider the linear system 3 *' = 27 a. Find the eigenvalues and eigenvectors for the coefficient matrix. help (numbers) help (matrices) A1 = , 01 = and help (numbers) help (matrices) 12 = , v2 = Find the real-valued solution to the initial value problem x = -3x1 – 2x2, x = 5x1+3x2, x1(0) = -4, x2(0) = 10. Use t as the independent variable in your answers. help (formulas) T2(t) help (formulas) OK Learn more Cookies help us deliver our services. By using our services, you agree to our use of cookies.

4. Solve given that x' = Ax + A has eigenvalues -2 and -6. -3 3 1 5

Find two linearly independent solutions of y" + 5xy = 0 of the form y1 = 1+ a3a + a6x® + ... y2 = x + b,x* + byz7 +... Enter the first few coefficients: az = a6 b4 by = || ||

3) Solve 1−sin2(A)+1 = 2cos(A) over [0,2π)