Using Variable Separable Method MCQ Quiz - Objective Question with Answer for Using Variable Separable Method - Download Free PDF

Calculation: 

⇒ 

Let 

∴ 

⇒ 

y(1) = 3

Hence, the correct answer is Option 1. 

Concept:

First Order Differential Equation:

• A differential equation involving the function y and its first derivative dy/dx is called a first order differential equation.
• Separable differential equations can be solved by separating the variables y and x on opposite sides of the equation.
• Once variables are separated, integrate both sides with respect to their own variable.
• Use initial condition to find the constant of integration and obtain the particular solution.

Logarithmic Function:

• The natural logarithm function is denoted as log_e or ln.
• Important identity: ∫(1/x) dx = log_e|x| + C

Calculation:

Given, y(0) = 1

Equation: (y − x²y) dy = (1 − x³) dx

⇒ y(1 − x²) dy = (1 − x³) dx

⇒ y dy = [(1 − x³)/(1 − x²)] dx

⇒ y dy = [(1 + x + x²)/(1 + x)] dx

⇒ y dy = x + (1/(1 + x)) dx

Integrate both sides,

⇒ ∫ y dy = ∫ (x + 1/(1 + x)) dx

⇒ y²/2 = x²/2 + log_e|1 + x| + C

⇒ y² = x² + 2 log_e|1 + x| + C′

Apply initial condition: x = 0, y = 1

⇒ (1)² = 0 + 2 log_e(1) + C′

⇒ 1 = 0 + 0 + C′

⇒ C′ = 1

∴ Hence, the particular solution is y² = x² + 2 log_e|1 + x| + 1

Calculation

⇒ 

⇒ 

⇒ 

⇒ 

⇒ 

= 2(ℓn2)² – 1 

Hence option 2 is correct

Calculation

Given:  . 

:⇒

Integrate both sides:

⇒

⇒

⇒

⇒

Given that when :

⇒

⇒

⇒

Substitute into the equation:

⇒

⇒

⇒

Hence option 1 is correct.

Calculation

Since, 

⇒ 

After integrating on both sides, we have

log(y + 1) = log(x - 1) - log C

C(y + 1) = (x - 1)

C = 

If x = 1, then y = 2, so C = 0.

Therefore, x - 1 = 0

Hence, there is only one solution.

Hence option 2 is correct

Concept:

Calculation:

Given: dy = (1 + y²) dx

Integrating both sides, we get

⇒ y = tan (x + c)

∴ The solution of the given differential equation is y = tan (x + c).

Calculation:

Given: 

On integrating both sides, we get

⇒ y = xe^a + c

Concept:

Calculation:

Given: 

Integrating both sides, we get

Concept: 

Calculation: 

Given :  

⇒  

Integrating both sides, we get 

⇒  

⇒ 

⇒   [∵ 2c = C]

⇒ 

The correct option is 2 . 

Concept:

Some useful formulas are:

If log x = z then we can write x = e^z

Calculation:

Rearranging the equation and integrating we get,

⇒

⇒, c = constant of integration

⇒ log(3x + 8) = 3(t + c)

⇒ 3x + 8 = e^3(t+c) 

⇒ 3x = e^3(t+c) - 8

∴ 

Concept:

Calculation:

Given: dy =  dx 

⇒  

Integrating both sides, we get

⇒  

⇒  = x + c 

⇒ y = sin ( x + c ) . 

The correct option is 2.

Calculation:

Given: ​

⇒ ydy = (x + 1) dx

Integrating both sides, we get

⇒ ∫ ydy = ∫ (x + 1) dx

⇒ 

⇒ y² = x² + 2x + 2c

∴ y2 - x2 - 2x - c = 0

Please note: c is constant here, so 2c can be also considered as a constant. 

Concept:

Calculation:

Given: 

Integrating both sides, we get

Concept:

For first-order differential equation, separate the variable and integrate accordingly.

Put the given condition to find out the integration constant

Calculation:

Given differential equation 

⇒ 

Integrating both sides

⇒ 

⇒ ln y = 4x + c

⇒ y = e^{4x + c}

Now y(0) = 1

⇒ 1 = e^{0 + c}

⇒ c = 0

∴ y = e^4x

Concept:Differential Equations by Variable Separable Method

If the coefficient of  is only function of x and coefficient of  is only a function of y in the given differential equation then we can separate both  and  terms and integrate both separately.

Calculation:

To Find: Solution of the differential equation

⇒ ydx - xdy = 0

⇒ ydx = xdy 

⇒ 

Integrating both sides, we get

                   (∵ ln x + ln y = ln (xy))

⇒ y = cx 

∴ y - cx = 0