Differential Equations - 2 - IIT JAM MCQ

we have (D³ - 9D)y = cos x
implies D(D² - 9)y = cos x


So, complete solution is

we have (D² + 6D + 25)y = 0

So, y =e^-3x [A cos 4x + B sin 4x]
but y(0) = 0 implies A = 0
Hence, y = Be^-3x sin 4x

we have xdy = (y + xe^-y/x)dx
implies xdy - ydx = xe^-y/x dx

Correct Answer :- A

Explanation : y''- m²y = 0

r² - m² = 0

r = +-m

y = Ae^mx + Be^-mx

Let A=C+D and B=D-C

= Ce^mx + De^mx - Ce^-mx + De^-mx

= C(e^mx-e^-mx) + D(e^mx+e^-mx)

= Csinh(mx) + Dcosh(mx)

Differentiating w.r.t. x, we get

Replacing  the equation of orthogonal trajectory is given by

implies 
Integrating both sides, we get

implies 2 tan^-1 3x + 3 In |y| = k

 is satisfied, So, equation is exacct
 ...(i)
 ...(ii)

Adding (i) and (ii), writing the common term once.

we have (D - 3)y₁ - 4y₂ = 0
4y₂ + (3 - D)y₁ = 0
Solving equations, we get

So, 

Taking total differential

we have


and 
Since, 
So, differential equation is exact
Now,

and

So, solution of differential equation

Multiplying the differential equation by 1/x², we get 
It is exact
So, 
implies 
2b + c = 0

we have y = In sin (x + a ) + b
implies y'= cot (x + a)
implies y" = - cosec² (x + a)

Degree of an equation is defined as the power of the highest derivative present in the equation. Hence from the equation, the degree is 5.

To determine which statement is true about the given differential equation, let's analyze each option:

a) The differential equation is linear:
A linear differential equation is one in which the dependent variable and its derivatives appear only to the first power and are not multiplied together. In the given equation, both x and y appear to the first power, but they are multiplied together in the terms (xy)dx and (2xy)dy. Therefore, the given differential equation is not linear.

b) The differential equation is exact:
A differential equation is exact if it can be written in the form M(x, y)dx + N(x, y)dy = 0, where M and N are functions of x and y, and ∂M/∂y = ∂N/∂x. Let's check if this condition holds for the given equation:
M(x, y) = x + y
N(x, y) = 2x + 2y
∂M/∂y = 1
∂N/∂x = 2
Since ∂M/∂y is not equal to ∂N/∂x, the given differential equation is not exact.

c) ex^y is an integrating factor of the differential equation:
An integrating factor is a function that can be multiplied to a differential equation to make it exact. In this case, the integrating factor would need to be able to transform the given equation into an exact one. However, as we determined in the previous option, the given equation is not exact. Therefore, ex^y is not an integrating factor.

d) A suitable substitution transforms the differential equation to the variable separable form:
Variable separable form means that the equation can be written in the form f(x)dx = g(y)dy, where f(x) and g(y) are functions of x and y, respectively. To determine if a suitable substitution can transform the given equation into this form, let's make the substitution x = u and y = v, where u and v are functions of a new variable t (u = u(t) and v = v(t)). Substituting these into the given equation, we have:
(uv)du + (2uv)dv = 0
Factoring out uv, we get:
uv(du + 2dv) = 0
Since this equation can be separated into f(u)du = -2g(v)dv, where f(u) = uv and g(v) = -v, we can conclude that a suitable substitution transforms the differential equation to the variable separable form.

Therefore, the correct answer is option d) A suitable substitution transforms the differential equation to the variable separable form.

(x - h)² + (y - k)² = a² ...(i)
where h, k are parameters 
Differentiating w.r.t. x, we get 
( x - h ) + ( y - k ) y' = 0 ...(ii)
Differentiating w.r.t. x again 1 + ( y - k)
y¹¹ + y¹²  = 0
implies   (iii)
From (ii), we get ( x - h ) = - ( y - k)y'
 ...(iv)
Putting (x - h) and (y - k) in (i), we get