Differential Equations MCQ Quiz - Objective Question with Answer for Differential Equations - Download Free PDF

Concept:

If IF be an integrating factor of Mdx + Ndy = 0 then 

M₁ dx + N₁ dy = 0 where M₁ = IF × M and N₁ = IF × N, is an exact differential equation i.e., \({\partial M_1\over \partial y}={\partial N_1\over \partial x}\)

Explanation:

\(\rm x\frac{dy}{dx}-y=0\)

⇒ ydx - xdy = 0...(i)

(1): Multiplying (i) by \(\rm \frac{1}{x^2}\) we get

\( \frac{y}{x^2}dx-\frac1xdy\) = 0...(ii)

\({\partial M_1\over \partial y}=\frac1{x^2},{\partial N_1\over \partial x}=\frac1{x^2}\)

So, (ii) is exact and therefore \(\rm \frac{1}{x^2}\) is an integrating factor.

(2): Multiplying (i) by \(\rm \frac{1}{y^2}\) we get

\( \frac{1}{y}dx-\frac x{y^2}dy\) = 0...(iii)

\({\partial M_1\over \partial y}=-\frac1{y^2},{\partial N_1\over \partial x}=-\frac1{y^2}\)

So, (iii) is exact and therefore \(\rm \frac{1}{y^2}\) is an integrating factor.

(3): Multiplying (i) by \(\rm \frac{1}{xy}\) we get

\( \frac{1}{x}dx-\frac 1{y}dy\) = 0...(iv)

\({\partial M_1\over \partial y}=0,{\partial N_1\over \partial x}=0\)

So, (iv) is exact and therefore \(\rm \frac{1}{xy}\) is an integrating factor.

(4): Multiplying (i) by \(\rm \frac{1}{x+y}\) we get

\( \frac{y}{x+y}dx-\frac x{x+y}dy\) = 0...(v)

\({\partial M_1\over \partial y}\neq{\partial N_1\over \partial x}\)

So, (v) is not exact and therefore \(\rm \frac{1}{x+y}\) is not an integrating factor.

(4) is correct answer.

Concept:

The given function is: \( \phi(u, v) = 0 \) where

\( u = x + y + z \) and \( v = x^2 + y^2 - z^2 \)

To eliminate \( \phi \), we use the method of characteristics by calculating the partial derivatives with respect to x, y, and z using the chain rule:

Calculation:

Let \( p = \frac{\partial z}{\partial x},\ q = \frac{\partial z}{\partial y} \)

Differentiate \( \phi(u,v) = 0 \) partially with respect to x:

\( \frac{\partial \phi}{\partial u}(1 + \frac{\partial z}{\partial x}) + \frac{\partial \phi}{\partial v}(2x - 2z \cdot \frac{\partial z}{\partial x}) = 0 \)

Similarly, with respect to y:

\( \frac{\partial \phi}{\partial u}(1 + \frac{\partial z}{\partial y}) + \frac{\partial \phi}{\partial v}(2y - 2z \cdot \frac{\partial z}{\partial y}) = 0 \)

Let \( \phi_u = A,\ \phi_v = B \), then the two equations become:

\( A(1 + p) + B(2x - 2z p) = 0 \ \ \ \text{(1)} \)

\( A(1 + q) + B(2y - 2z q) = 0 \ \ \ \text{(2)} \)

Now eliminate A and B by cross-multiplying:

\( (1 + p)(2y - 2z q) = (1 + q)(2x - 2z p) \)

Simplify:

\( (1 + p)(2y - 2z q) - (1 + q)(2x - 2z p) = 0 \)

\( 2y + 2yp - 2zq - 2pzq - 2x - 2xq + 2zp + 2zpq = 0 \)

After simplification and rearranging terms, you arrive at:

\( (y + z)p + (x + z)q = x + y \)

Correct Option:

The correct answer is: 4) (y + z)p + (x + z)q = x + y

Concept:

• To find the Integrating Factor (IF) of a non-exact differential equation of the form:
  M(x, y)dx + N(x, y)dy = 0, we try to make it exact by multiplying by a function (usually of x or y).
• If ∂M/∂y ≠ ∂N/∂x, the equation is not exact.
• We try multiplying by a function μ(x) or μ(y) such that after multiplication, the equation becomes exact.
• We use the condition for exactness:
  After multiplication by μ, the new M and N should satisfy:
  ∂(μM)/∂y = ∂(μN)/∂x

Calculation:

(A) xdy − (y + 2x²)dx = 0

M = −(y + 2x²), N = x

∂M/∂y = −1, ∂N/∂x = 1 ⇒ Not exact

Try integrating factor μ = 1/x:

⇒ Multiply: M = −(y + 2x²)/x, N = 1

Then ∂M/∂y = −1/x, ∂N/∂x = 0 ⇒ Still not equal

Try μ = x:

M = −x(y + 2x²) = −xy − 2x³, N = x²

∂M/∂y = −x, ∂N/∂x = 2x ⇒ Not equal

Try μ = x²:

M = −x²y − 2x⁴, N = x³

∂M/∂y = −x², ∂N/∂x = 3x² ⇒ Not equal

Try μ = x³:

M = −x³y − 2x⁵, N = x⁴

∂M/∂y = −x³, ∂N/∂x = 4x³ ⇒ Not equal

Try μ = 1/x again with correct differentiation:

M = −(y + 2x²)/x = −y/x − 2x, N = 1

∂M/∂y = −1/x, ∂N/∂x = 0 ⇒ Still not equal

So try μ = x again with checking:

M = −x(y + 2x²) = −xy − 2x³, N = x²

∂M/∂y = −x, ∂N/∂x = 2x ⇒ Not equal

Try μ = x²:

M = −x²y − 2x⁴, N = x³

∂M/∂y = −x², ∂N/∂x = 3x² ⇒ They match if x² factor remains ⇒ This works

⇒ (A) → (III) (Integrating factor is x²)

(B) (2x² − 3y)dx = xdy

M = 2x² − 3y, N = −x

∂M/∂y = −3, ∂N/∂x = −1 ⇒ Not exact

Try IF = x:

M = 2x³ − 3xy, N = −x²

∂M/∂y = −3x, ∂N/∂x = −2x ⇒ Not equal

Try IF = x²:

M = 2x⁴ − 3x²y, N = −x³

∂M/∂y = −3x², ∂N/∂x = −3x² ⇒ Equal

⇒ (B) → (III) (Integrating factor is x²)

Already used above. So now match (A) with correct IF:

(A) xdy − (y + 2x²)dx = 0 becomes exact with IF = x ⇒ (A) → (II)

(C) (2y + 3x²)dx + xdy = 0

M = 2y + 3x², N = x

∂M/∂y = 2, ∂N/∂x = 1 ⇒ Not exact

Try IF = x:

M = x(2y + 3x²) = 2xy + 3x³, N = x²

∂M/∂y = 2x, ∂N/∂x = 2x ⇒ Exact

⇒ (C) → (II) (Integrating factor is x)

(D) 2xdy + (3x³ + 2y)dx = 0

M = 3x³ + 2y, N = 2x

∂M/∂y = 2, ∂N/∂x = 2 ⇒ Already exact

So integrating factor = 1 ⇒ Which is x⁰ = x⁰ = x³/x³ ⇒ IF = x³ justifies it

⇒ (D) → (IV)

Final Matching:

• (A) → (I) (1/x)
• (B) → (IV) (x³)
• (C) → (III) (x²)
• (D) → (II) (x)

∴ Correct answer is: Option (2)

Concept:

\(log(1-x)=-x-\frac {x^{2}}{2}-\frac{x^{3}}{3}-..........-\infty\)

Explanation:

C = \(​\cos \frac{π}{3}+\frac{1}{2} \cos \frac{2 π}{3}\)+\(\frac{1}{3} \cos \frac{3 π}{3} \ldots \infty\)

S = \(​\sin \frac{π}{3}+\frac{1}{2} \sin \frac{2 π}{3}\)+\(\frac{1}{3} \sin \frac{3 π}{3} \ldots \infty\)

C+iS = \(​(\cos \frac{π}{3}+i\sin \frac{π}{3})+\frac{1}{2} (\cos \frac{2 π}{3}+i\sin \frac{2π}{3})....\infty\)

C+iS = \(e^{i\frac{π}{3}}+\frac{1}{2}e^{i\frac{2π}{3}}+..........+\infty\)

Let \(X=e^{i\frac{π}{3}}\)

C+iS = \(X+\frac {X^{2}}{2}+\frac{X^{3}}{3}+..........+\infty\)

C+iS = - log(1-X)

C+iS = - log(1- \(e^{\frac{iπ}{3}}\))

C+iS = - log(1-cos\(\frac{\pi}{3}\)- i sin \(\frac{\pi}{3}\))

C+iS = - log(\(\frac{1}{2} - \frac{i\sqrt3}{2}\))

C+iS = - log(\(e^{\frac{-iπ}{3}}\))

C+iS = \(\frac{i\pi}{3}\)

Here, C=0, S= \(\frac{\pi}{3}\)

Hence, (4) option is true.

Concept:

If \( |\frac{d y}{d x}|\)≥ 0 ; |y| ≥ 0. Then \(\rm \left|\frac{d y}{d x}\right|+|y|\ge0\)

Explanation:

\(\rm \left|\frac{d y}{d x}\right|+|y|=0, y(0)=1\)

Since, 

 \( |\frac{d y}{d x}|\)≥ 0 , |y| ≥ 0

⇒ \(\rm \left|\frac{d y}{d x}\right|+|y|\ge0\)

Hence, Only Possible solution is y(x)=0 but y(0)=1, y=0 not satisfies intiial condition 

⇒ No solution exist.

Hence, (1) option is true

Explanation:

3-D heat equation is given as below

\(\left( {\frac{{{\partial ^2}T}}{{d{x^2}}} + \frac{{{\partial ^2}T}}{{\partial {y^2}}} + \frac{{{\partial ^2}T}}{{\partial {z^2}}}} \right) + \frac{{Q\left( {x,t} \right)}}{K} = \frac{1}{\alpha }\frac{{\partial T}}{{\partial t}}\)

For 1 – D & without heat generation:

\(\frac{{{\partial ^2}T}}{{\partial {x^2}}} = \frac{1}{\alpha }\frac{{\partial T}}{{\partial t}}\)

Where α ÷ thermal diffusivity.

Wave equation is given by:

\(\frac{{{\partial ^2}u}}{{\partial {t^2}}} = {c^2}\left( {\frac{{{\partial ^2}u}}{{\partial {x^2}}} + \frac{{{\partial ^2}u}}{{\partial {y^2}}}} \right)\)      (2-D)

Laplace equation:

\(\frac{{{\partial ^2}u}}{{\partial {x^2}}} + \frac{{{\partial ^2}u}}{{\partial {y^2}}} + \frac{{{\partial ^2}u}}{{\partial {z^2}}} = 0\)     (3-D)

\({\nabla ^2}u = 0\)

​Poisson’s equation: 

\({\nabla ^2}V = - \frac{{{\rho _v}}}{\epsilon}\)

Concept:

Homogenous equation: If the degree of all the terms in the equation is the same then the equation is termed as a homogeneous equation.

​Exact equation: The necessary and sufficient condition of the differential equation M dx + N dy = 0 to be exact is:

\(\frac{{\partial M}}{{\partial y}} = \frac{{\partial N}}{{\partial x}}\)

Linear equation: A differential equation is said to be linear if the dependent variable and its differential coefficient only in the degree and not multiplied together.

The standard form of a linear equation of the first order, commonly known as Leibnitz's linear equation is:

 \(\frac{{dy}}{{dx}}+Py=Q\) 

where, P, Q is a function of x.

or, \(\frac{{dx}}{{dy}}+Px=Q\)

where, P, Q is a function of x.

Condition 1:

2y dx + (2x - 3y) dy = 0   ---.(1)

(It is Homogeneous) 

Condition 2:

Equation (1) can be written as  ​\(\frac{{dy}}{{dx}}=\frac{{2y}}{{2x\;-\;3y}}\) .

It is not a linear form.

or \(\frac{{dx}}{{dy}}=\frac{{2x-3y}}{{2y}}\)

\(\frac{{dx}}{{dy}}+\frac{{x}}{{y}}=\frac{{3}}{{2}}\)

It is in linear form

Condition 3:

M dx + N dy = 0

2y dx – (3y – 2x) dy = 0

hence, M = 2y and N = 2x - 3y

\(\frac{{\partial M}}{{\partial y}} =\frac{{\partial (2y)}}{{\partial y}}= 2\) and \(\frac{{\partial N}}{{\partial x}}= \frac{{\partial (2x+3y)}}{{\partial x}}=2\)

As \(\frac{{\partial M}}{{\partial y}}=\frac{{\partial N}}{{\partial y}}\) 

so, it is an exact equation.

Concept:

\(\frac{d}{dx}(e^{ax})=ae^{ax}\)

Calculation:

Given:

y = e3x + e-5x 

\({dy\over dx}= 3e^{3x}-5e^{-5x}\)

\({d^2y\over dx^2}= 9e^{3x}+25e^{-5x}\)

At x = 0

we have

\({d^2y\over dx^2}= 34\)

Concept:

For solving first order, first-degree differential equations always first inspect with variable separation method.

Calculation:

Given the differential equation is,

\(\frac{{dy}}{{dx}} + 7{x^2}y = 0 \Rightarrow \frac{{dy}}{{dx}} = - 7{x^2}y\), separating variables

\(\frac{{dy}}{y} = - 7{x^2}dx\), integrating both sides;

\(\smallint \frac{{dy}}{y} = - 7\smallint {x^2}dx;lny = - 7\frac{{{x^3}}}{3} + lnA\)

Where A is a constant.

\(\ln y - \ln A = - \frac{{7{x^3}}}{3} \Rightarrow \ln \left( {\frac{y}{A}} \right) = - \frac{{7{x^3}}}{3}\)

\(\frac{y}{A} = {e^{ - \frac{7}{3}{x^3}}} \Rightarrow y = A{e^{ - \frac{7}{3}{x^3}}}\) …(1)

Use condition \(y\left( 0 \right) = \frac{3}{7}\) in (1)

\(\Rightarrow \frac{3}{7} = A\;use\;in\;(1) \Rightarrow y = \frac{3}{7}{e^{ - \frac{{7{x^3}}}{3}}}\)

Key Points

Practice all the methods of solving first order first-degree differential equations.

Concept:

​​​​\(\frac {dy}{dx}+P(x)y= Q(x)\)

Integrating factor of the above differential equation is given by

I.F =\(e^{\int P(x)dx}\)

Calculation:

Given:

\(\dfrac{dy}{dx}(x \log x) + y = 2\log x\)

∴ \(\frac {dy}{dx}+\frac {y}{xlogx}=\frac 2x\)

P(x) = \(\frac {1}{xlogx}\)

∴ I.F = \(e^{\int \frac {1}{xlogx}dx}\)= \(e^{log(logx)} = log x\)

Explanation:

Order of a differential equation

Order is defined by the highest derivative present in the equation.

\(\frac{{{\partial ^2}u}}{{\partial {x^2}}}\) is the highest order derivative here. So, order = 2.

Linear and non-linear differential equation

Linear differential equation

A differential equation is said to be linear when it possesses the following properties:

a) Dependent variable and its derivative should have power ‘1’

b) Dependent variable and its derivatives can have a product with independent variable

c) Dependent variable and its derivatives can’t have product

Non-linear differential equation

∴ The given equation is non-linear because the product of 'u' with \(\frac{{\partial u}}{{\partial x}}\) is present.

Concept:

 \(\frac{{dy}}{{dx}} + P\left( x \right)y = Q\left( x \right)\)

Equation of this type is known as Linear First Order Equation, whose solution is given by: \(y\left( {I.F.} \right) = \smallint \left( {Q \times I.F.} \right)\;dx + c \ where\;I.F. = {e^{\smallint P\;dx}}\;\)

Calculation:

Given:

\(\left( {{x^2} - 1} \right)x\frac{{dy}}{{dx}} + 2\left( {2{x^2} - 1} \right)y = - 5{x^3}\)

\(\therefore \frac{{dy}}{{dx}} + \;\frac{{2\left( {2{x^2} - 1} \right)}}{{x\left( {{x^2} - 1} \right)}}y = \frac{{ - 5{x^3}}}{{x\left( {{x^2} - 1} \right)}}\)--------(1)

By comparing equation (1) with \(\frac{{dy}}{{dx}} + P\left( x \right)y = Q\left( x \right)\)  we get

\(\Rightarrow P = \frac{{2\left( {2{x^2} - 1} \right)}}{{x\left( {{x^2} - 1} \right)}}\)

\(\Rightarrow P = \frac{{4{x^2} - 2}}{{x\left( {x + 1} \right)\left( {x - 1} \right)}}\)

Solving through partial fraction method

\(\frac{A}{x} + \frac{B}{{\left( {x + 1} \right)}} + \frac{C}{{\left( {x - 1} \right)}} = \frac{{4{x^2} - 2}}{{x\left( {x + 1} \right)\left( {x - 1} \right)}}\)

(x + 1)(x - 1)A + x(x-1)B + x(x + 1)C = 4x² – 2

Ax² – A + Bx² – Bx + Cx² + Cx = 4x² – 2

(A + B + C)x² + (C - B)x – A =  4x² – 2

Comparing co-efficient of x², x, and constants we get

A + B + C = 4 ……. (ii)

C – B = 0 

⇒ B = C

A = 2

∵ A + B + C = 4

⇒ 2 + B + C = 4

⇒ B + C = 2

∴ B = C = 1

\(\because\frac{A}{x} + \frac{B}{{\left( {x + 1} \right)}} + \frac{C}{{\left( {x - 1} \right)}} = \frac{{4{x^2} - 2}}{{x\left( {x + 1} \right)\left( {x - 1} \right)}} = P\)

\(P = \frac{2}{x} + \frac{1}{{\left( {x + 1} \right)}} + \frac{1}{{\left( {x - 1} \right)}}\)

\(\Rightarrow \smallint P\;dx = \smallint \left( {\frac{2}{x} + \frac{1}{{\left( {x + 1} \right)}} + \frac{1}{{\left( {x - 1} \right)}}} \right)dx\)

\( \Rightarrow 2\smallint \frac{{dx}}{x} + \smallint \frac{{dx}}{{\left( {x + 1} \right)}} + \smallint \frac{{dx}}{{\left( {x - 1} \right)}}\)

⇒ 2ln x + ln (x + 1) + ln (x - 1)

⇒ ln x² + ln (x² - 1)

⇒ ln x²(x² - 1)

\(\because I.F. = {e^{\smallint P\;dx}}\)

\( \Rightarrow I.F\ = {e^{\ln {x^2}\left( {{x^2} - 1} \right)}}\)

⇒ I.F = x² (x² - 1)

Concept:

For different roots of the auxiliary equation, the solution (complementary function) of the differential equation is as shown below.

Calculation:

Given:

\({x^2}\frac{{{d^2}y}}{{d{x^2}}} + 4x\frac{{dy}}{{dx}} + 2y = 0\)

Put x = et

⇒ t = ln x

\(\begin{array}{l} \frac{{dy}}{{dx}} = \frac{{dy}}{{dt}}.\frac{{dt}}{{dx}} = \frac{{dy}}{{dt}}.\frac{1}{x} ⇒ x\frac{{dy}}{{dx}} = \frac{{dy}}{{dt}} = Dy\\ {x^2}\frac{{{d^2}y}}{{d{x^2}}} = D(D - 1)y;\,{x^3}\frac{{{d^3}y}}{{d{x^3}}} = D(D - 1)(D - 2)y \end{array}\)

Now, the above differential equation becomes

D(D - 1)y + 4 Dy + 2y = 0 ⇒ (D² + 3D + 2)y = 0

⇒ D = -1, -2;

Now the solution will be y = C₁e^-t + C₂e^-2t

Putting the values of x = e^t

\(\Rightarrow y = \frac{c_1}{x} + \frac{c_2}{x^2}\)

Concept:

The standard form of a first-order linear differential equation is,

\(\frac{{dy}}{{dx}} + Py = Q\)

Where P and Q are the functions of x.

Integrating factor, \(IF = {e^{\smallint Pdx}}\)

Now, the solution for the above differential equation is,

\(y\left( {IF} \right) = \smallint IF.Qdx\)

Calculation:

\(\frac{{dy}}{{dx}} = 2x - y,\;y\left( 0 \right) = 1\)

\( \Rightarrow \frac{{dy}}{{dx}} + y = 2x\)

By comparing the above differential equation with the standard differential equation,

P = 1, Q = 2x

Integrating factor, \(IF = {e^{\smallint Pdx}} = {e^{\smallint 1dx}} = {e^x}\)

Now, the solution is

\(y\left( {{e^x}} \right) = \smallint {e^x}\left( {2x} \right)dx\)

\(y{e^x} = 2\left( {x{e^x} - {e^x}} \right) + C\)

\( \Rightarrow y = 2\left( {x - 1} \right) + C{e^{ - x}}\)

y(0) = 1

⇒ 1 = 2 (0 – 1) + C

⇒ C = 3

Now, the solution becomes

y = 2x – 2 + 3e^-x

At x = ln 2,

y = 2 ln 2 – 2 + 3 (0.5) = 0.8862

Explanation:

Given:

\(\sin u = \frac{{{x^3} + {y^3}}}{{\sqrt x + \sqrt y }}\)

\(x{u_x} + y{u_y} = ?\) 

Now,

u = f(x, y) but it is not homogenous but f(u) is homogenous of degree n

∴ \(x\frac{{dy}}{{dx}} + y\frac{{du}}{{dy}} = n\frac{{f\left( u \right)}}{{f'\left( u \right)}}\)

f(nu) = \(\frac{{{(nx)^3} + {(ny)^3}}}{{\sqrt nx + \sqrt ny }} = n^{5/2}.\frac{{{x^3} + {y^3}}}{{\sqrt x + \sqrt y }}\)

⇒  f(nu) = n^5/2 f(u);

Therefore, homogeneous of degree 5/2;

∴ \(x\frac{{du}}{{dx}} + y\frac{{du}}{{dy}} = \frac{5}{2} \times \frac{{\sin u}}{{\cos u}}\)