Department of Mathematics, IIT Madras
MA 2020 - Differential Equations, Instructor: Prof. A. K. B. Chand
2022-2023 : First Semester - BATCH-1
Problem Sheet-3: Second Order Linear Homogeneous/Non-homogeneous Differential Equations
1. Show that y1 = e−x and y2 = e2x are solutions of the equation y 00 − y 0 − 2y = 0. Also find the general
solution.
2. Show that each of the functions y1 (x) = sin x − sin 3x
3
and y2 (x) = sin3 x is a solution of the differ-
ential equation y 00 + (tan x − 2 cot x)y 0 = 0. Further, verify that y = c1 y1 + c2 y2 is not the general
solution of the differential equation.
3. Find the general solution of the following equations where one solution y1 is given:
(a) x2 y 00 + xy 0 − y = 0, y1 = x; (b) y 00 + y = 0, y1 = sin x;
(c) (1 − x)2 y 00 − 2xy 0 + 2y = 0, y1 = x; (d) x2 y 00 + xy 0 + (x2 − 14 )y = 0, y1 = x(−1/2) sin x;
(e) xy 00 − (2x + 1)y 0 + (x + 1)y = 0, y1 = ex ; (f) )x2 y 00 + xy 0 − 4y = 0; y1 = x2
4. Solve the following:
(a) y 00 + 3y 0 − 54y = 0; (b) y 00 + 4y 0 + 4y = 0; (c) y 00 − 6y, +13y = 0;
(d) y 00 − 9y, +20y = 0; (e) 16y 00 − 8y 0 + y = 0; (f ) y 00 = 4y;
(g) y 00 + 4y = 0, y(0) = 3 and y 0 (0) = 8; (h) y 00 − 5y 0 + 6y = 0, y(1) = e2 and y 0 (1) = 3e2 ;
(i) y 00 + 8y 0 − 9y = 0, y(1) = 2 and y 0 (1) = 0; (j) y 00 + 4y 0 + 5y = 0, y(0) = 1 and y 0 (0) = 0.
5. Solve the following problems by using the method of variation of parameters:
e−x ex
(a) y 00 + 2y 0 + y = x2
; (b) y 00 + a2 y = sec ax; (c) y 00 − 3y 0 + 2y = 1+ex
;
(d) y 00 − 2y 0 − 3y = 64xe−x ; (e) y 00 + 2y 0 + y = e−x log x; (f) 2y 00 + 3y 0 + y = −3x
;
(g) y 00 − 4y 0 + 3y = (1 + e−x )−1 ; (h) y 00 + y = sec x tan x; (i) x2 y 00 − 4xy 0 + 6y = 21x−4 ;
(j) x2 y 00 − 2xy 0 + 2y = x3 cos x; (k)xy 00 − y 0 = (3 + x)x2 ex ;
(l) y 00 + x1 y 0 − 1
x2
y = log x, given that y1 = x and y2 = 1/x are two linearly independent solutions of
its reduced equation;
(m) (x2 + x)y 00 + (2 − x2 )y 0 − (2 + x)y = x(x + 1)2 , if the complementary function is c1 ex + c2 /x.
6. Verify that y = c1 x−1 + c2 x5 is a solution of x2 y 00 − 3xy 0 − 5y = 0, on any interval [a, b] that does
not contain the origin.
7. Consider the differential equation x2 y 00 + xy 0 − 4y = 0.
(a) Show that x2 and 1/x2 are linearly independent solutions of it on the interval 0 < x < ∞.
(b) Write the general solution of the given equation.
(c) Find the solution that satisfies the conditions y(2) = 3, y 0 (2) = −1. Explain why this solution is
8. Given that x, x2 and x4 are all solutions of x3 y 000 − 4x2 y 00 + 8xy 0 − 8y = 0. Also show that they are
linearly independent on 0 < x < ∞ and write the general solution.
9. Define the Wronskian W (y1 , y2 ) of any two differentiable functions y1 and y2 defined in an interval
(a, b) ⊂ R. Show that W (y1 , y2 ) = 0 if y1 and y2 are linearly dependent.
10. If y1 and y2 are any two solutions of a second order linear homogeneous ordinary differential equation
which is defined in an interval (a, b) ⊂ R, then W (y1 , y2 ) is either identically zero or non-zero at any
point of the interval (a, b).
11. Consider the two functions f (x) = x3 and g(x) = x2 |x| on the interval [−1, 1].
(a) Show that their Wronskian W (f, g) vanishes identically.
(b) Show that f and g are not linearly dependent.
12. Find a second order homogeneous linear ODE for which the given functions are solutions. Show
linear independence by the Wronskian. Solve the IVP.
(a) 1, e−2x ; y(0) = 1, y 0 (0) = −1
(b) e−kx cos πx, e−kx sin πx, y(0) = 1, y 0 (0) = −k − π
13. Solve, using the method of undetermined coefficients:
(a) y 00 − 3y 0 + 2y = 14 sin 2x − 18 cos 2x; (b) y 00 − 4y 0 + y = x2 − 2x + 2;
(c) y 00 + 9y = 2 sin 3x + 4 sin x − 26e−2x + 27x3 ; (d) y 00 − 2y 0 + 2y = ex sin x;
(e) y 00 − 2y 0 + y = xex ; (f) y 00 + 10y 0 + 25y = 14e−5x ;
(g) y 00 − 3y 0 = x + ex sin x (h) y 00 + 2y = ex + 2.
dn
14. Find the general solution of each of the following equations: (Note Dn ≡ dxn
).
(a) (D3 − 4D2 + 5D − 2)y = 0;
(b) (D2 − 3D + 2)y = (4x + 5)e3x ;
(c) (D2 − 2D + 3)y = 3e−x cos x;
(d) (D3 − 2D2 − 5D + 6)y = 18ex ;
(e) (D2 + 25)y = 50 cos 5x + 30 sin 5x;
(f) (D3 − 2D2 + 4D − 8)y = 8(x2 + cos 2x);
(g) (D3 + 3D2 − 4)y = 12e−2x + 9ex .
