Differential Equations - 1 - Free MCQ Test with solutions for Maths

we have dy/dx = 2y + z and ...(i)
dz/dx = 3y  ...(ii)
from 3 (i) l (ii), we get 

implies 
on integrating, we get

implies ln(3y + z) = 3x + c
So, 3y + z = ke^3x

This special case where D has a co-efficient is solved using Legendre’s method. 

Let log⁡(8x + 7) = z 

Then, e^z = 8x + 7

Substituting this in the equation, 

8Dy + 2y = x 

y(8D + 2) = x 

Thus the auxiliary equation is 8D + 2 = 0 

Thus, D = -1 / 4

(Solving by substituting powers of e^z)

The General solution is C.F + P.I =

1. Option D is correct.

   M(x,y) = 2y and N(x,y) = 2x - 3y.

   Compute the partial derivatives: ∂M/∂y = 2 and ∂N/∂x = 2. Since these are equal, the differential form is exact.

   Both M and N are homogeneous of degree 1 (each term scales linearly with a common factor), so the equation is homogeneous.

   Rearrange to express dx/dy: dx/dy = (3y - 2x)/(2y) = 3/2 - x/y. This can be written as dx/dy + (1/y)·x = 3/2, which is a first-order linear equation in the dependent variable x with independent variable y.

   Therefore the equation is exact, homogeneous, and also linear when viewed as an ODE for x(y); hence option D is correct.

2(cos(y²)) dx - xy sin(y²) dy  

∂M/∂y = -2 sin(y²) ⋅ (2y)  
       = -4y sin(y²)  

∂N/∂x = -y sin(y²)  

(∂M/∂y - ∂N/∂x) / N = (+3y sin(y²)) / (+x y sin(y²)) = 3/x  

∴ If I = e^(∫(3/x) dx) = x³.
 

There will only one constant in the first-order differential equation. Differentiating the given equation.

 

Putting the value of c in Eq. (1) and simplifying we will get a first-order and second-degree equation. Hence, (A) is the correct answer.

we have 

Let z = log x, Then

The differential equation changes to

implies

So, 

=> x² ln⁡x dy - xy dy=xy dx – y² ln⁡y dx …….dividing by x² y² then


on integrating we get

where c is a constant of integration.

y₃(x) and y₄(x) are linearly dependent if y₃(x) = ky₄(x) where k is any constant 

Since, y₁(x) and y₂(x) are independent
So, a - kc = 0
and b - kd = 0
or 
So, y₃(x) and y₄(x) are independent if

implies ad ≠ bc

we have 
or, 
Thus, 
implies y = x² + c for x ≤ 0,
but y(0) = 0 implies c = 0
So,
 

we have (D² + D + 3)y = 5 cos(2x + 3)
So,

Let z = log x, Then 

So, The differential equation becomes

So, 

The order of an equation is defined as the highest order derivative present in the equation and hence from the equation,it is clear that it of 2^nd order.

we have 
implies 

So, Equation becomes,

Differentiating (x - 1)² + y² + 2ax = 0 with respect to x gives

2(x - 1) + 2y(dy/dx) + 2a = 0.

From the family equation, 2ax = -[(x - 1)² + y²], so

a = -((x - 1)² + y²)/(2x).

Substituting a into the differentiated equation and simplifying yields

x² - y² - 1 + 2xy(dy/dx) = 0.

For orthogonal trajectories, replace the slope of the given family, dy/dx, by the negative reciprocal of the slope of the orthogonal curves. Let dy/dx denote the derivative for the orthogonal trajectories; then substitute dy/dx_of_family = -1/(dy/dx).

Replacing and solving for dy/dx gives

dy/dx = 2xy / (x² - y² - 1).

Thus the differential equation of the orthogonal trajectories is dy/dx = 2xy / (x² - y² - 1), which corresponds to option A.

Given equation, 

Dividing both sides by 2x, we get 

There is a linear differential equation of the form: 

The integrating factor is found by:

Multiply both sides of the equation by the integrating factor:

The solution  represents a family of parabolas. This is because the equation has the form of a quadratic curve (which describes a parabola). 

Therefore, the answer is c) parabolas

Given, 

Taking the natural logarithm of both sides to eliminate the exponential term:

The order of a differential equation is the highest derivative present in the equation.

Here, the highest derivative is y′′y''y′′ (second derivative), so the order of the equation is 2.

The degree of a differential equation is the exponent of the highest derivative, provided that the equation is polynomial in the derivatives.

In this case, the equation llny=y′+y′′ is not polynomial in y′ and y′′ due to the presence of the logarithmic function. Therefore, the degree is 1.

The order is 2, and the degree is 1.

Thus, the sum of the degree and order is 2+1=3

(a) is the correct answer.