Equation Of First Order
And
First Degree
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Separation of variables
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If in an equation, it is possible to get all the functions of x and dx in to one side and all the function of y and dy to the other, the variable are said to be separable.
Working rule to solve an equation in which variables are separable.
step 1 )
Let dy/dx =f₁(x) f₂(y) ..(1)
be given equation f₁(x) is a function of x alone and f₂(y) is a function of y alone.
step 2)
From (1) separating variables,
[1/f₂(y)] dy = f₁(x) dx ............(2)
step 3)
Integrating both sides of (2), we have ∫[1/f₂(y)] dy=∫f₁(x)dx + c...(3)
where c is constant of integration, is the required solution.
Note 1:
In all solution (3), an arbitrary constant c must be added in any one side only. If c is not added, then the solution obtained will not be a general solution of (1).
Note 2: To simplify the solution (3), the constant of integration can be chosen in any suitable form so as to get the final solution in a form as simple as possible. Accordingly, we are write log c, tan⁻¹ c, sin c, eᶜ,
(1/2). C , (-1/3). C etc in place of c in some solutions.
Note 3 :
The students are advised to remember by heart the following formulas. These will help them to write solution (3) in compact form
i) log x+ log y= log xy
ii) log x - log y = log(x/y)
iii) n log x = log xⁿ
iv)tan⁻¹x+tan⁻¹y= tan⁻¹[(x+y)/(1-xy)]
v) tan⁻¹x-tan⁻¹y = tan⁻¹[(x-y)/(1+xy)]
vi) eˡᵒᵍ ᶠ⁽ˣ⁾ = f(x).
EXAMPLE
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1) dy/dx = eˣ⁻ʸ + x² e⁻ʸ
dy/dx = e⁻ʸ( eˣ +x²)
or eʸ dy = (x² +eˣ) dx
integrating eʸ = x³/3 + eˣ + c, c is con.
2) √(1+x²+y²+x²y²) + xy (dy/dx) = 0
=√{(1+x²)(1+y²)} + xy(dy/dx) = 0
= √(1+x²)dx /x + ydy/√(1+y²) =0
=(1+x²)dx/x√(1+x²)+ydy/√(1+y²)
=0
= ∫ dx/x√(1+x²) +∫xdx/√(1+x²)+
ydy/(1+y²) =c
= log x - log{1- √(1+x²)}+
√(1+x²)+√(1+y²) =c
EXERCISE
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1) dy/dx = eˣ⁺ʸ + x² eʸ
2) (dy/dx)tan y=sin(x + y) + sin(x-y)
3) dy/dx=(sinx+xcosx)/
{y(2log y+1)}
4) dy/dx={x(2log x+1)}/
(siny +ycosy)
5) log(dy/dx) = ax + by
6) y - x(dy/dx) = a(y² + dy/dx)
7) 3eˣ tan y dx + (1- eˣ)sec² y dy =0
8) ₑx+y dy ₌ ₓ² ₑx³+y dx
9) dy/dx = eˣ ⁺ ʸ when x=1, y=1. find y when x= -1
10) (eˣ + 1)y0 dy = (y +1)eˣ dx
11) (dy/dx) - y tan x = - y sec² x
12) x√(1+y²) dx + y√(1+x²) dy =0
13) (2ax+x²)(dy/dx) = a² + 2ax
14) dr = a (r sinθ dθ - cosθ dr)
15) (eʸ +1) cosx dx + eʸsin x dy = 0
16) √(a+x) (dy/dx) +x = 0
17) dy/dx = √{(1-y²)/(1 -x²)}
18) (x²-yx²)dy + (y²+xy²) dx =0
19) (xy² +x)dx + (yx² +y) dy = 0
20) sec²x tany dx+sec²y tan xdy =0
21) (1+x)y dx + (1+y)x dx = 0
22) (1- x²)(1 - y) dx = xy(1+y)dx
23) x²(y+1)dx + y²(x - 1) dy =0
24) (dy/dx)tan y=sin(x+y)+sin(x - y)
25) y - x dy/dx = 3(1+ x² dy/dx)
26) cosy log(secx +tanx) dx =
cos xlog(sec y + tan y) dy
27) x dy - y dx = (a² + y²)¹/² dx
