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Differential Equations solved questions
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CHAPTER 46Differential Equations
46.1 Solve
The variables are separable: 7y dy = 5x dx. Then, J ly dy = J 5x dx,
Here, as usual, C represents an arbitrary constant.
46.2 Solve
The variables are separable. 7 In |y| =51n|jt| + C I ,
s c
e 5 ( in M+c,)/7 = gCl. e scin Mjn |y| = C2\x\ '\ where C2 = e ' >0. Note that any function of the form y=
Cc5'7 satisfies the given equation, where C is an arbitrary constant (not necessarily positive).
46.3 Solve
Separate the variables. In | y | = kx + C,, \ y \ = ek*+c> = eCl • e1" = C2ekx. Any function
of the form y = Cekx is a solution.
46.4 Solve
Note that y = Ce" '2 is a general
solution.
46.5 Solve
Separate the variables.
Taking reciprocals, we get
Hence, If we allow C to be negative (which means allowing Cl to be complex),
46.6 Solve
46.7 Solve
46.8 Solve
46.9 Solve
431
, 432 CHAPTER 46
46.10 Solve
46.11 Solve
tan""1 y = tan~' x + C,,
where C = tan C,.
46.12 Solve
e3y = 3ln\l + x\ + C, 3y = In(3In |l + *| + C), y=
46.13 Solve
sin ' y = sin ' x + C,
y = sin (sin ' x + C) = sin (sin * AC) cos C + cos (sin * x) sin C = x cos C + Vl — x2 sin C
46.14 Solve given
Since C = -2. Hence,
46.15 Solve
46.16 Solve x dx + y dy = xX* dy — y dx).
46.17 Solve / = (* + y)2. As usual
[Tie variables do not separate. Try the substitution z = x + y. Then, z' = l + y'. So z' -1 = z2,
Now, the variables x and z are separable. tan ' z = x + C, z = tan (x + C),
* + >> = tan (x + C), y = tan (A: + C) - x.
46.18 Solve
The variables are not separable. However, on the right side, the numerator and denominator are
homogeneous of the same degree (one) in x and y. In such a case, let y = vx. Then,
Hence, Thus, Now the variables x and
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