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Edurev123
9.1 Solve the following initial value DEs :
???? ?? ''
+?? ?? '
+?? =?? ,?? '
(?? )=?? .?? and ?? '
(?? )=?? .
(2017 : 7 Marks)
Solution:
Auxiliary equation using Euler Cauchy equation
20?? 2
+4?? +1=0
?? =
-4±v16-80
40
=
-4±8?? 40
=
-1
10
±
?? 5
?? =?? -?? /10
[?? cos
?? 5
+?? sin
?? 5
]
?? (0) =3.2??? =3.2
?? (?? ) =?? -
?? 10
[-
4
5
sin
?? 5
+
?? 5
cos
?? 5
]-
1
10
?? -
?? 10
[?? cos
?? 5
+?? sin
?? 5
]
?? (0) =
?? 5
-
?? 10
=0
?? =
?? 2
=1.6
? ?? (?? ) =?? -
?? 10
[3.2cos
?? 5
+1.6sin
?? 5
]
9.2 Solve the initial value problem:
?? -?? ?? +?? ?? =?? ?? ?? ?? (?? ) =
????
????
,?? (?? )=
?? ??
(2018 : 13 Marks)
Solution:
Given equation can be written as
?? 2
?? -5???? +4?? =?? 2?? (?? 2
-5?? +4)?? =?? 2??
Page 2
Edurev123
9.1 Solve the following initial value DEs :
???? ?? ''
+?? ?? '
+?? =?? ,?? '
(?? )=?? .?? and ?? '
(?? )=?? .
(2017 : 7 Marks)
Solution:
Auxiliary equation using Euler Cauchy equation
20?? 2
+4?? +1=0
?? =
-4±v16-80
40
=
-4±8?? 40
=
-1
10
±
?? 5
?? =?? -?? /10
[?? cos
?? 5
+?? sin
?? 5
]
?? (0) =3.2??? =3.2
?? (?? ) =?? -
?? 10
[-
4
5
sin
?? 5
+
?? 5
cos
?? 5
]-
1
10
?? -
?? 10
[?? cos
?? 5
+?? sin
?? 5
]
?? (0) =
?? 5
-
?? 10
=0
?? =
?? 2
=1.6
? ?? (?? ) =?? -
?? 10
[3.2cos
?? 5
+1.6sin
?? 5
]
9.2 Solve the initial value problem:
?? -?? ?? +?? ?? =?? ?? ?? ?? (?? ) =
????
????
,?? (?? )=
?? ??
(2018 : 13 Marks)
Solution:
Given equation can be written as
?? 2
?? -5???? +4?? =?? 2?? (?? 2
-5?? +4)?? =?? 2??
C.F. : The auxiliary equation is
?? 2
-5?? +4 =0?(?? -4)(?? -1)=0
?? =1 or ?? =4
?? ?? =?? 1
?? ?? +?? 2
?? 4?? , where ?? 1
and ?? 2
are constants.
P.I.:
?? ?? =
1
?? 2
-?? +4
?? 2?? =
1
(?? -4)(?? -1)
?? 2?? =
1
(?? -4)
×
1
1
×?? 2?? =
-?? 2?? 2
? ?? =?? ?? +?? ?? =?? 1
?? ?? +?? 2
?? 4?? -
?? 2?? 2
Given:
?? (0)=
19
12
=?? 1
+?? 2
-
1
2
?? 1
+?? 2
=
25
12
(??)
?? (0) =
8
3
??? 1
?? 0
+?? 2
×4?? 0
-?? 0
=
8
3
? ?? 1
+4?? 2
=
11
3
(???? )
Solving (i) and (ii), we get
?? 2
=
19
36
;?? 1
=
14
9
? ?? =
14
9
?? ?? +
19
36
?? 4?? -
?? 2?? 2
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FAQs on Application to Initial Value Problem for 2nd Order Linear Equation with Constant Coefficient - Mathematics Optional Notes for UPSC
| 1. How do you determine the initial conditions for a 2nd order linear equation with constant coefficients? |  |
Ans. To determine the initial conditions for a 2nd order linear equation with constant coefficients, you need to be given the values of the dependent variable and its derivative at a specific initial time, usually denoted as t=0.
| 2. What is the general solution to a 2nd order linear equation with constant coefficients? | |
Ans. The general solution to a 2nd order linear equation with constant coefficients is a linear combination of two functions, typically exponential functions, sine and cosine functions, or a combination of these, depending on the roots of the characteristic equation.
| 3. How do you apply the initial value problem to solve a 2nd order linear equation with constant coefficients? | |
Ans. To solve the initial value problem for a 2nd order linear equation with constant coefficients, you first find the general solution to the equation. Then, apply the initial conditions given to determine the specific values of the constants in the general solution.
| 4. Can the initial value problem for a 2nd order linear equation with constant coefficients have multiple solutions? | |
Ans. No, the initial value problem for a 2nd order linear equation with constant coefficients typically has a unique solution once the initial conditions are specified.
| 5. What role do the initial conditions play in solving the initial value problem for a 2nd order linear equation with constant coefficients? | |
Ans. The initial conditions determine the specific solution to the 2nd order linear equation with constant coefficients. They provide the necessary information to find the values of the constants in the general solution and uniquely determine the solution that satisfies both the equation and the initial conditions.
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Nov 21, 2024 Last updated |
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