Math, asked by JeffreyDs, 7 months ago

find dy/dx if y =[x²+(x³+Sin2x)]....its from differential calculus pls explain ​

Answers

Answered by Legend42
11

Answer:

both m1 and m2 are constants, the expressions (D – m1) (D – m2) y and (D– m2) (D – m1) y are equivalent i.e. the expression is independent of the order of operational factors.

We will explain the method with the help of following

Example: Solve d2y/ dx2 – 5 dy/dx +6y = e3x.

Solution:

The equation can be written as

(D2 – 5D + 6)y = e3x

(D – 3) (D – 2)y = e3x

C.F. = c1 e3x + c2e2x

And P.I. =1/ (D-3). 1/ (D-2) e3x

= 1/ (D-3) e2x ∫ e3x e-2xdx

=1/ (D-3) e2x ex

= e3x ∫ e3x e-3x dx = x.e3x

y = c1e3x +c2e2x +xe3x

P.I. can be found by resolving

1/f (D) = 1/ (D-3). 1/ (D-2)

Now using partial fractions,

1/f (D) = 1/ (D-3). 1/ (D-2)

= 1/ (D-3) – 1/ (D-2)

Hence, the required P.I. is [1(D-3) – 1/ (D-2)] e3x

= 1/ (D-3) e3x – 1/ (D-2) e3x

= e3x ∫ e3x e-3x dx – e2x ∫ e3x e-2x dx

= xe3x – e3x

Second term can be neglected as it is included in the first term of the C.F.

Short Method of Finding P.I.

In certain cases, the P.I. can be obtained by methods shorter than the general method.

(i). To find P.I. when X = eax in f(D) y = X, where a is constant

y = 1/f(D)

1/f(D) eax = 1/f(a) eax , if f(a) ≠ 0.

1/f(D) eax = xr/f r(a) eax , if f(a) = 0, where f(D) = (D-a)rf(D)

Example: Solve (D3 – 5D2 + 7D – 3)y = e3x.

Solution:

(D – 1)2 (D – 3) y = e3x

C.F. = aex + bx ex + ce3x

And P.I. = 1/ (D3-5D2+7D-3) e3x

= x/ (3D2-10D+7) e3x

= ¼ xe3x

y = aex + bx ex + ce3x + ¼ xe3x.

(ii). To find P.I. when X = cos ax or sin ax

f (D) y = X

y = 1/f(D) sinax

If f (– a2) ≠ 0 then 1/f(D2) sinax = 1/f(-a2) sinax

If f (– a2) = 0 then (D2 + a2) is at least one factor of f (D2)

Let f (D2) = (D2 + a2)r f (D2)

Where f (– a2) ≠ 0

[1/f (D2)] sin ax = 1/ (D2+a2). 1/φ(D2) sin ax

= 1/ φ(-a2). 1/ (D2+a2) sin ax

Now, when r=1, 1/(D2 + a2) sin ax = -x/2a. cos ax

Similarly If f(–a)2 ≠ 0 then 1/f(D2) cos ax = 1/f(-a2) cosax

And 1/ (D2+a2) cos ax = x/2a .sin ax

Example: Solve (D2 – 5D + 6) y = sin3x.

Solution: We can factorize the given term as (D – 2) (D – 3)y = sin3x

C.F. = ae2x + be3x

P.I. = 1/ (D2 -5D +6) sin 3x

= 1/ [(-9) -5D +6] sin 3x

= 1/ (-5D-3) sin 3x

= - (5D-3). 1/ (25D2 -9) sin3x

= 1/234 (15 cos 3x-3 sin3x)

= ae2x + be3x + 1/234 (15 cos 3x-3 sin3x)

(iii) To find the P.I. when X = xm where m ∈ N

f (D) y = xm

y = 1/ f(D) xm

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