Differential equations Comprehensive Questions (Frequently Tested) and Complete Solutions Graded A+

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uB Questions and Answers Sheet 10
Differential equations
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Question #1
Find the general solution to the following differential equations.

𝑦 ′′ + 6𝑦 ′ + 13𝑦 = 0

Answer:
Given: 𝑦 ′′ + 6𝑦 ′ + 13𝑦 = 0
𝑦 ′′ + 6𝑦 ′ + 13𝑦 = 0
A second order linear, homogenous ODE has the form ay''+by'+cy=0
For an equation ay''+by'+cy=0 , assume a solution of the form 𝑒 𝜆𝑡
Rewrite the equation with 𝑦 = 𝑒 𝜆𝑡
′′ ′
(𝑒 𝜆𝑡 ) + 6(𝑒 𝜆𝑡 ) + 13𝑒 𝜆𝑡 = 0
𝑒 𝜆𝑡 (𝑦 2 + 6𝑦 + 13) = 0
𝑦 2 + 6𝑦 + 13 = 0
Roots of the quadratic equation in the form ax^{2}+bx+c=0 is
−𝑏 ± √𝑏 2 − 4𝑎𝑐
𝑥1,2 =
2𝑎
Here, 𝑎 = 1, 𝑏 = 6 and 𝑐 = 13
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−6 ± √62 − 4 × 1 × 13
𝑦1,2 =
2×1
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−6 ± √36 − 52
𝑦1,2 =
2
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𝑦1,2 = −3 − 2𝑖, −3 + 2𝑖
For two complex roots 𝑦1 ≠ 𝑦2 , where 𝑦1 = −3 − 2𝑖, 𝑦2 = −3 + 2𝑖




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So, the general solution takes the form: 𝑦 = 𝑒 𝛼𝑡 (𝑐1 cos(𝛽𝑡) + 𝑐2 sin(𝛽𝑡))
𝑦 = 𝑒 −3𝑡 (𝑐1 cos(2𝑡) + 𝑐2 sin(2𝑡))

Question #2
Find the general solutions of the differential equations 6𝑦 4 + 11𝑦 ′′ + 4𝑦 = 0

Answer:
6𝑦 4 + 11𝑦 ′′ + 4𝑦 = 0(1)
Eq (1) can be written as
(6𝐷4 + 11𝐷2 + 4)𝑦 = 0
AE. 6𝑚4 + 11𝑚2 + 4 = 0
For value of m we put 𝑚2 = 𝑢
𝑚 4 = 42
6𝑢2 + 11𝑢 + 4 = 0
𝑢 = −1/2 𝑢 = −4/3
𝑚2 = −1/2𝑚2 = −4/3
2√3 2
𝑚 = ±𝑖√1/2𝑚 = ±𝑖√4/3 = ±𝑖 = ±𝑖
3 √3
Which is comple roots.
1 1 2𝑡 2𝑡
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𝑦 = 𝑐1 cos (√ 𝑡) + 𝑐2 sin (√ 𝑡) + 𝑐3 cos ( ) + 𝑐4 sin ( )
2 2 √3 √3
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Question #3
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Determine whether the following differential equations are exact or not exact. if exact, solve for the
uB general solution

(𝑥 + 2𝑦)𝑑𝑥 − (2𝑥 + 𝑦)𝑑𝑦 = 0
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Answer:
Step 1

(𝑥 + 2𝑦)𝑑𝑥 − (2𝑥 + 𝑦)𝑑𝑦 = 0
The differential equation Mdx+Ndy=0 is a exact if
𝜕𝑀 𝜕𝑁
=
𝜕𝑦 𝜕𝑥

Step 2

From the given differential equation
𝑀 = 𝑥 + 2𝑦
𝑁 = −(2𝑥 + 𝑦)
𝑁 = −(2𝑥 + 𝑦)
𝜕𝑀
=2
𝜕𝑦
𝜕𝑁
= −2
𝜕𝑥
𝜕𝑀 𝜕𝑁
Since, ≠
𝜕𝑦 𝜕𝑥
The given differential equation is not exact differential equation.
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Question #4
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Evaluate the following differential equations:
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Integrating Factor by Formula




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(2𝑦 2 + 2𝑦 + 4𝑥 2 )𝑑𝑥 + (2𝑥𝑦 + 𝑥)𝑑𝑦 = 0

Answer:
𝑀𝑑𝑥 + 𝑁𝑑𝑦 = 0
𝜕𝑀 𝜕𝑁
= 4𝑦 + 2 = 2𝑦 + 1
𝜕𝑦 𝜕𝑥
𝜕𝑀 𝜕𝑁
≠ 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛 𝑖𝑠 𝑛𝑜𝑡 𝑒𝑥𝑎𝑐𝑡.
𝜕𝑦 𝜕𝑥
𝑀𝑦 − 𝑁𝑥 (4𝑦 + 2) − (2𝑦 + 1)
𝐼𝐹 = =
𝑁 2𝑥𝑦 + 𝑥
4𝑦 − 2𝑦 + 2 − 1 2𝑦 + 1 1
= = =
𝑥(2𝑦 + 1) 𝑥(2𝑦 + 1) 𝑥
𝑀𝑦 − 𝑁𝑥 1
𝐼𝐹 = = = 𝑓(𝑥)
𝑁 𝑥
1
Finally if for given D.E is =
𝑥


Question #5
Find the general solution of the given differential equations

4𝑦 ′′ − 4𝑦 ′ − 3𝑦 = 0
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Answer:
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