Differential Equations solved questions

CHAPTER 46
Differential 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