Question

3. Find the general solutions for the following homogeneous ODEs. dºy.dy + y = 0 a) dx2 dx d²y b) dx2 4y = 0 a) d²y dy + dx² dx = 0

Answer 1

dy t a) O z dy t da dne anniliary equation m + mti zo solving for me -lt Wi-4x1x1 z (for quadratic equation an tont(20, na-b+ √²-hac 29 ✓-3 -17 V-3 m 2 3 XI 2 23 m 2 -1 + 3i 2 mz m -1 - 31 2 -1t3i 2 2 "

In this case ศ 24 CF 2 [C, cos Bx + Q singa CF or complementary junction mi m, 2-Y ti when 2 tip M 22-iß 2 here ti 3 m 2 - - 13/ 2 2 >> L2-1/2 B2 3½ -/n hence CF e C, cos(3n) + q sin(3/27 PI 0 Sine the hand side right os the equation doesn't have terms init any 'n'containing teams init dy + dy ty=0 / doesnot have any team du ty= containing 'n hence PI=0

Now solution ac CF e PI 2. -1/2 [C, cos ( 722 ) + sin (½ 2 e ar م) 2 dy dx2 . 2 ay auxiliary equation 2 4 O z 2 4 mz +2 22 +2 m 2. M₂ An uns case + m₂ n Coz tma с. е the M2 m. 2+2 22 – 2 - 2n (F 2 C, en + S

since the / in containing here abo PI 20 Egnation does not have terms' in it. any Now solution yz CF+PI -an ye gent qe c O 2 dyt dy da? dn egration auxiliary 2 mt m = 2 2 M mao auel mizam ) mz -1 hence mhas a values (6, and -1) - m m 20 2 2 min CF2 M₂n ciem

On - 1xn CF e C et se -n C, Xit t ce PI 20, reason as for the same 2 Questions above. solution ac 2 CFT PI cita ento -a ce z Senti please leave I like if you are happy with hi answer,