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

Concept:

Differential Equation and Curve Passing Through a Point:

• If the slope of a tangent to a curve at any point is known, we use dy/dx to represent it.
• We can form a differential equation if a relationship is given between x, y, and dy/dx.
• Solving this differential equation gives the equation of the curve.
• If the curve passes through a specific point, we use that point to find the constant of integration.
• We substitute the given options into the final curve equation to find the correct point through which it also passes.

Calculation:

Given,

dy/dx = y / x

This is because slope × x = y

⇒ (dy/dx) × x = y

⇒ dy/dx = y / x

⇒ dy / y = dx / x

Integrating both sides,

⇒ ∫(1 / y) dy = ∫(1 / x) dx

⇒ ln |y| = ln |x| + C

⇒ ln |y| − ln |x| = C

⇒ ln |y / x| = C

⇒ y / x = A (where A = e^C)

⇒ y = A × x

Now apply point (1, 1) to find A:

⇒ 1 = A × 1 ⇒ A = 1

⇒ So curve is y = x

Check which point lies on y = x

(A) (−1, 2): 2 ≠ −1 ⇒ Incorrect

(B) (√3, 0): 0 ≠ √3 ⇒ Incorrect

(C) (2, 2): 2 = 2 ⇒ Correct

(D) (3, 0): 0 ≠ 3 ⇒ Incorrect

∴ The correct point is (2, 2).

Concept:

The equation of the parabola having a vertex at the (h,k) and axis along the positive y-axis is (x - h)² = 4a(y - k)

Calculation:

The equation of the family of parabolas having vertex at origin and axis along the positive y-axis

⇒ (x - 0)² = 4a(y - 0) 

⇒ x² = 4ay  __(i)

Differentiating both sides with respect to x,

⇒ 2x = 4a\(\frac{\text{dy}}{\text{dx}}\)

⇒ x = 2a \(\frac{\text{dy}}{\text{dx}}\) __(ii)

Putting the value of a from (i) in (ii),

⇒ x = 2\(x^2 \over 4y\) \(\frac{\text{dy}}{\text{dx}}\)

⇒ 2xy = x²\(\frac{\text{dy}}{\text{dx}}\)

⇒ x\(\frac{\text{dy}}{\text{dx}}\) − 2y = 0

∴ The correct option is (2).

Concept:

The slope of normal of a curve = \({-dx \over dy}\)

Area of right-angle triangle = \({1 \over 2}\) × (perpendicular) × (base)

Calculation:

Equation of normal to the curve at (x,y) 

is (Y - y) = -\({dx \over dy}\)(X - x)

Let the normal to the curve cut the x and y-axis at A(a,0) and B(0,b) respectively.

⇒ A(a,0) and B(0,b) satisfies the equation of normal

⇒ (0 - y) = -\({dx \over dy}\)(a - x) and (b - y) = -\({dx \over dy}\)(0 - x)

⇒ a = y\(dy \over dx\) + x and b = x\({dx \over dy}\) + y __(i)

Now, Area( triangle OAB) = 1 {where O(0,0) is the origin}

⇒ \({1 \over 2}\)OA × OB = 1

⇒ \({1 \over 2}\)a × b = 1

⇒ \({1 \over 2}\)× (y\(dy \over dx\) + x )(x\({dx \over dy}\) + y) = 1

⇒ (xy + y²\(dy \over dx\)  + x²\({dx \over dy}\) + xy) = 2

Multiply the equation with \(dy \over dx\) both sides,

⇒ 2xy\(dy \over dx\) + y2(\(dy \over dx\))² + x2 = 2\(dy \over dx\)

⇒  y2(\(dy \over dx\))2 + 2\(dy \over dx\)(xy - 1) + x2 = 0

∴ The correct answer is option (4).

Concept:

Slope of tangent to curve = dy/dx

Calculation:

Given: the slope of the tangent at any point (x, y) is equal to the product of its co-ordinates

\(\rm \frac{d y}{d x}=x y\\\ \rm \frac{1}{y} d y=x d x\\ Integrate, \log y=\frac{x^{2}}{2}+C\ldots(1)\\ Equation (1) passing \ through (1,1)\\ \log (1)=\frac{1^{2}}{2}+C\\ \Rightarrow C=-\frac{1}{2}\\ Put \ in \ eq.(1) \Rightarrow \log y=\frac{x^{2}}{2}-\frac{1}{2}\\ \quad \Rightarrow \quad 2 \log y=x^{2}-1\)

Concept:

In ∫x dy, x will be considered as constant

In ∫x dy, x will be considered as constant

Formula Used:

∫xⁿdx = xⁿ⁺¹/(n+1)

Calculation:

Here, x(dx - dy) + y(dy - dx) = 0

⇒ xdx - x dy + ydy - ydx = 0

Taking integration, we get 

∫ xdx - ∫ x dy + ∫ ydy - ∫ ydx = 0

By using the above formula 

⇒ \(\rm \frac{x^2}{2}-xy+\frac{y^2}{2}-xy+c=0\\ \)

⇒ \(\rm x^2+y^2-4xy+c=0\)

\(\Rightarrow \rm x^2+y^2 =4xy+c\)

Hence, option (2) is correct. 

Concept:

In ∫x dy, x will be considered as constant

In ∫x dy, x will be considered as constant

Formula Used:

∫xⁿdx = xⁿ⁺¹/(n+1)

Calculation:

Here, x(dx - dy) + y(dy - dx) = 0

⇒ xdx - x dy + ydy - ydx = 0

Taking integration, we get 

∫ xdx - ∫ x dy + ∫ ydy - ∫ ydx = 0

By using the above formula 

⇒ \(\rm \frac{x^2}{2}-xy+\frac{y^2}{2}-xy+c=0\\ \)

⇒ \(\rm x^2+y^2-4xy+c=0\)

\(\Rightarrow \rm x^2+y^2 =4xy+c\)

Hence, option (2) is correct. 

Concept:

Slope of tangent to curve = dy/dx

Calculation:

Given: the slope of the tangent at any point (x, y) is equal to the product of its co-ordinates

\(\rm \frac{d y}{d x}=x y\\\ \rm \frac{1}{y} d y=x d x\\ Integrate, \log y=\frac{x^{2}}{2}+C\ldots(1)\\ Equation (1) passing \ through (1,1)\\ \log (1)=\frac{1^{2}}{2}+C\\ \Rightarrow C=-\frac{1}{2}\\ Put \ in \ eq.(1) \Rightarrow \log y=\frac{x^{2}}{2}-\frac{1}{2}\\ \quad \Rightarrow \quad 2 \log y=x^{2}-1\)

Concept:

Variable Separable Form:

If it is possible to write a differential equation in the form of f(x) dx = g(y) dy where, f(x) is a function of x and g(y) is a function of y, then we say that the variables are separable.

This kind of differential equations can be solved by integrating both sides. The solution is given by \(\int f\left( x \right)~dx=~\int g\left( y \right)~dy+C\)

Calculation:

\(\frac{dx}{dt}~\propto x\)        ---(Given)

\(\Rightarrow \frac{dx}{dt}=k~x\)

As it is given that, proportional constant equal to 2

\(\Rightarrow \frac{dx}{dt}=2x\)

\(\Rightarrow \frac{dx}{2x}=dt\)

Now, by integrating both the sides we get

\(\Rightarrow ~\int \frac{dx}{2x}=~\int dt\)

\(\Rightarrow \ln \sqrt{x}=~t+{{C}_{1}}\)

Now, raising both the side to the power of e, we get

Concept:

The equation of the parabola having a vertex at the (h,k) and axis along the positive y-axis is (x - h)² = 4a(y - k)

Calculation:

The equation of the family of parabolas having vertex at origin and axis along the positive y-axis

⇒ (x - 0)² = 4a(y - 0) 

⇒ x² = 4ay  __(i)

Differentiating both sides with respect to x,

⇒ 2x = 4a\(\frac{\text{dy}}{\text{dx}}\)

⇒ x = 2a \(\frac{\text{dy}}{\text{dx}}\) __(ii)

Putting the value of a from (i) in (ii),

⇒ x = 2\(x^2 \over 4y\) \(\frac{\text{dy}}{\text{dx}}\)

⇒ 2xy = x²\(\frac{\text{dy}}{\text{dx}}\)

⇒ x\(\frac{\text{dy}}{\text{dx}}\) − 2y = 0

∴ The correct option is (2).

Calculation:

dR/dt = KR

dR = KR dt

\(\smallint \frac{{dR}}{R}\; = \;\smallint k\;dt\)

In (R) = Kt + C1

R = ekt + c1 = ekt.ec1 = cekt

For point (0, 100)

100 = ce°, c = 300

For point (1, 300)

R = 100ekt, 300 = 100ek.1

ek = 300/100 = 3

k = In (300/100) = ln 3

R = 100ekt = 100 (ek)t

R = 100 (ek)t

R = 100 (3)t

To determine the half-life of a radioisotope is to use the fraction left undecayed as a measure of how many half-lives have passed. The mass of a radioactive isotope gets halved with the passing of every half-life, which means that

• 1 half-life→1/2 left undecayed

• 2 half-lives→1/4 left undecayed

• 3 half-lives→1/8 left undecayed

• 4 half-lives→1/16 left undecayed

Therefore, 4 half-lives must pass until you have 1/16 of the original sample. Mathematically, this means that

t / t_1/2 = 4

Since 2 hours have passed i.e. 120 minutes have passed

 Therefore, the half-life of the substance is 

 120 / t_1/2 = 4 or t_1/2 = 120 / 4 =30 min

Concept:

Let the variable is x which is varying with time t.

\(\frac{dx}{dt} \propto x\)

Calculation:

Given:

\(\frac{dx}{dt} \propto x\)

\(⇒ \frac{dx}{dt} =kx\)

\(⇒ \int\frac{dx}{x} =\int kdt\)

\(⇒ ln x =kt + p\)

Where p is a constant.

Now, take antilog,

⇒ x = e^{(kt +p)}

⇒ x = e(kt) e^p

Let c = ep

⇒ x = c e(kt) 

So, the variation will be exponential.

Let x² + y² = t

Differentiating with respect to x, we get:

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

\(\frac{y}{x}\frac{{dy}}{{dx}} = \frac{1}{{2x}}\frac{{dt}}{{dx}} - 1\) 

Hence, the given equation becomes:

\(\frac{1}{{2x}}\frac{{dt}}{{dx}} - 1 + \frac{{2t - 1}}{{t + 1}} = 0\) 

\(\frac{1}{{2x}}\frac{{dt}}{{dx}} = 1 - \frac{{2t - 1}}{{t + 1}}\) 

\(= \frac{{2 - t}}{{t + 1}}\) 

\(2xdx = \frac{{t + 1}}{{2 - t}}dt\) 

\(2xdx + \left( {1 + \frac{3}{{t - 2}}} \right)dt = 0\) 

Integrating, we have

x² + t + 3 log (t – 2) = C

2x² + y² + 3 log (x² + y² – 2) = C

Which is the required solution.

If x(t) is the amount of radium present in time t years.

According to the formula, \(\frac{{dx}}{{dt}} = kx\)

Hence, x = x₀ e^kt

If x₀ = 100, then x = 100 e^kt

Take logarithm on both sides we get,

\(t = \frac{{log\left( {\frac{x}{{100}}} \right)}}{k}\)

Also \(x\left( {1500} \right) = 50\;then,\: 50 = 100{e^{1500k}}\)

\(1500k = \log \frac{{50}}{{100}} = - \log \left( 2 \right)\;or,\)

\(k = \frac{{ - {\rm{log}}\left( 2 \right)}}{{1500}} = 0.0002\)

When x = 50,

\(t = \frac{{log\left( {\frac{x}{{100}}} \right)}}{k}\)

\(t = \frac{{log\left( {\frac{{50}}{{100}}} \right)}}{{\frac{{- {\rm{log}}\left( 2 \right)}}{{1500}}}}\)

\(t = - 1500 \times \frac{{log\left( {\frac{1}{2}} \right)}}{{\log \left( 2 \right)}}\)

\(t = - 1500 \times \frac{{ - 0.301}}{{0.301}}\)

t = 1500 Years approximately.

Concept:

In ∫x dy, x will be considered as constant

In ∫x dy, x will be considered as constant

Formula Used:

∫xⁿdx = xⁿ⁺¹/(n+1)

Calculation:

Here, x(dx - dy) + y(dy - dx) = 0

⇒ xdx - x dy + ydy - ydx = 0

Taking integration, we get 

∫ xdx - ∫ x dy + ∫ ydy - ∫ ydx = 0

By using the above formula 

⇒ \(\rm \frac{x^2}{2}-xy+\frac{y^2}{2}-xy+c=0\\ \)

⇒ \(\rm x^2+y^2-4xy+c=0\)

\(\Rightarrow \rm x^2+y^2 =4xy+c\)

Hence, option (2) is correct.