Differential Equations - 6 - Mathematics MCQ

Equation of circles touching y axis is given by 
( x - a²) + y² = a² where a is parameter implies x² + y² - 2ax = 0
Differentiating w.r.t. x, we get 
2x + 2yy' - 2a = 0 
implies a = x + yy'
So, differential equation will be 
x² +y²- 2x(x + yy¹) = 0
implies 
So, differential equation is non-linear and of first order 

we have 
on integrating, we get 
implies 
implies 
implies 
implies 
implies 
where c₁' is constant.

dy/dx + 1 = cosec (x + y)
Let x + y = t and 1 + dy/dx = dt/dx
⇒ dt/(cosec t) = dx
∴ ∫ sin t dt = ∫ dx
⇒ − cos t = x − c
⇒ cos(x + y) + x = c

we have (D² + 4D + 4)y = 0
implies D + 2)² y = 0
So, solution is 
but y(0) = 4 implies c₁ = 4
So, 

So, solution is given by (16x + 4) e^-2x

we have yy" - y'2 = 0
implies 
on integrating, we get 
implies In y' = In y + c
implies y' = c₁y
Again on integrating, 
implies In y = c₁x + k
implies 

we have (y² sin x + x cos y ) dx - ( x sin y - 2y sin x) dy = 0


So, differential equation is exact 

So, solution is given as

we have  
implies  ...(i)

implies  ...(ii)
Differentiating (i) w.r.t. x, we get 
 ...(iii)
Adding (ii) and (iii), we get 


and 
So, complete solution is given as

Thus, solution is

implies 

Here, C.F. = c₁ sinh x + c₂ cosh x


So, 
but y(0) = 0 implies c₂ = 0,
Thus 
implies 
At 

So, 

Here I.F. = 1/y², we get
Multiplying differential equation by 1/y²

Now, 
Since, 
So, equation becomes exact 

So, n = 4

Here, y'(x +y²) = y
implies 
implies 
So, 
Thus, solution is 
implies 
implies x = cy + y²

Here y' = 2^x-y
implies 2^y dy = 2^x dx
implies 
implies 2^x-2^y = c'

implies 
Let cot y= 2
Thus, 
implies 
So, 
Hence, solution is 
implies z = cot y = -x³ + cx²