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

Calculation: 

$\frac{\mathrm{d}\mathrm{y}}{\mathrm{d}\mathrm{x}}-\frac{\mathrm{y}}{\mathrm{x}}={\mathrm{y}}^3(1+{\log }_{\mathrm{e}}\mathrm{x})$

⇒ $\frac{1}{{\mathrm{y}}^3}\frac{\mathrm{d}\mathrm{y}}{\mathrm{d}\mathrm{x}}-\frac{1}{\mathrm{x}{\mathrm{y}}^2}=1+{\log }_{\mathrm{e}}\mathrm{x}$

Let $-\frac{1}{{\mathrm{y}}^2}=\mathrm{t}\Rightarrow \frac{2}{{\mathrm{y}}^3}\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{dt}}{\mathrm{dx}}$

∴ $\frac{\mathrm{d}\mathrm{t}}{\mathrm{d}\mathrm{x}}+\frac{2\mathrm{t}}{\mathrm{x}}=2(1+{\log }_{\mathrm{e}}\mathrm{x})$

$\text{ I.F. }={\mathrm{e}}^{\int \frac{2}{\mathrm{x}}\mathrm{dx}}={\mathrm{x}}^2$

⇒ $\frac{-{\mathrm{x}}^2}{{\mathrm{y}}^2}=\frac{2}{3}((1+{\log }_{\mathrm{e}}\mathrm{x}){\mathrm{x}}^3-\frac{{\mathrm{x}}^3}{3})+\mathrm{C}$

y(1) = 3

$\frac{{\mathrm{y}}^2}{9}=\frac{{\mathrm{x}}^2}{5-2{\mathrm{x}}^3(2+{\log }_{\mathrm{e}}{\mathrm{x}}^3)}$

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

$ydy=(xdy-ydx)\sin \left(\frac{x}{y}\right)$

⇒ $\frac{dy}{y}=\left(\frac{xdy-ydx}{y^2}\right)\sin \left(\frac{x}{y}\right)$

⇒ $\frac{\mathrm{dy}}{\mathrm{y}}=\sin \left(\frac{\mathrm{x}}{\mathrm{y}}\right)\mathrm{d}(-\frac{\mathrm{x}}{\mathrm{y}})$

⇒ $\ell ny=\cos \frac{x}{y}+C$

$x(1)=\frac{\pi }{2}\Rightarrow 0=\cos \frac{\pi }{2}+C\Rightarrow C=0$

⇒ $\ell \mathrm{ny}=\cos \frac{x}{y}$

$\text{ but }y=2\Rightarrow \cos \frac{x}{2}=\ln 2$

⇒ $\cos x=2{\cos }^2\frac{x}{2}-1$

= 2(ℓn2)² – 1 

Hence option 2 is correct

Calculation

Given:  $\frac{dy}{dx}=-4x{y}^2$. 

:⇒ $\frac{dy}{y^2}=-4xdx$

Integrate both sides:

⇒ $\int y^{-2}dy=\int -4xdx$

⇒ $\frac{y^{-1}}{-1}=-4\frac{x^2}{2}+C$

⇒ $-\frac{1}{y}=-2x^2+C$

⇒ $\frac{1}{y}=2x^2-C$

Given that $y=1$ when $x=0$:

⇒ $\frac{1}{1}=2(0{)}^2-C$

⇒ $1=-C$

⇒ $C=-1$

Substitute $C=-1$ into the equation:

⇒ $\frac{1}{y}=2x^2-(-1)$

⇒ $\frac{1}{y}=2x^2+1$

⇒ $y=\frac{1}{2x^2+1}$

Hence option 1 is correct.

Calculation

Since, $\frac{dy}{dx}=\frac{y+1}{x-1}$

⇒ $-\frac{dy}{y+1}=\frac{dx}{x-1}$

After integrating on both sides, we have

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

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

C = $\frac{x-1}{y+1}$

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:

${\displaystyle \int \frac{\mathrm{d}\mathrm{x}}{1+{\mathrm{x}}^2}={\tan }^{-1}\mathrm{x}+\mathrm{c}}$

Calculation:

Given: dy = (1 + y²) dx

$\Rightarrow \frac{\mathrm{d}\mathrm{y}}{1+{\mathrm{y}}^2}=\mathrm{d}\mathrm{x}$

Integrating both sides, we get

$$\begin{gathered} \Rightarrow {\displaystyle \int \frac{\mathrm{d}\mathrm{y}}{1+{\mathrm{y}}^2}={\displaystyle \int \mathrm{d}\mathrm{x}}} \\ \Rightarrow {\tan }^{-1}\mathrm{y}=\mathrm{x}+\mathrm{c} \end{gathered}$$

⇒ y = tan (x + c)

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

Calculation:

Given: $\ln \left(\frac{\mathrm{d}\mathrm{y}}{\mathrm{d}\mathrm{x}}\right)-\mathrm{a}=0$

$\Rightarrow \ln \left(\frac{\mathrm{d}\mathrm{y}}{\mathrm{d}\mathrm{x}}\right)=\mathrm{a}$

$\Rightarrow \frac{\mathrm{d}\mathrm{y}}{\mathrm{d}\mathrm{x}}={\mathrm{e}}^{\mathrm{a}}$

$\Rightarrow \int \frac{\mathrm{d}\mathrm{y}}{\mathrm{d}\mathrm{x}}=\int {\mathrm{e}}^{\mathrm{a}}$

On integrating both sides, we get

⇒ y = xe^a + c

Concept:

$\int \frac{1}{\mathrm{x}}\mathrm{d}\mathrm{x}=\log \mathrm{x}+\mathrm{c}$

$\int {\mathrm{x}}^{\mathrm{n}}\mathrm{d}\mathrm{x}=\frac{{\mathrm{x}}^{\mathrm{n}+1}}{\mathrm{n}+1}+\mathrm{c}$

Calculation:

Given: $(\mathrm{x}\mathrm{y}\frac{\mathrm{d}\mathrm{y}}{\mathrm{d}\mathrm{x}}-1)=0$

$\Rightarrow \mathrm{x}\mathrm{y}\frac{\mathrm{d}\mathrm{y}}{\mathrm{d}\mathrm{x}}=1$

$\Rightarrow \mathrm{y}\mathrm{d}\mathrm{y}=\frac{\mathrm{d}\mathrm{x}}{\mathrm{x}}$

Integrating both sides, we get

$\Rightarrow \frac{{\mathrm{y}}^2}{2}=\log \mathrm{x}+\mathrm{c}$

Concept: 

$\int \frac{1}{{\mathrm{a}}^2+{\mathrm{x}}^2}\mathrm{d}\mathrm{x}=\frac{1}{\mathrm{a}}{\tan }^{-1}\frac{\mathrm{x}}{\mathrm{a}}+\mathrm{C}$ 

Calculation: 

Given : $\mathrm{d}\mathrm{y}=(4+{\mathrm{y}}^2)\mathrm{d}\mathrm{x}$ 

⇒ $\frac{\mathrm{d}\mathrm{y}}{4+{\mathrm{y}}^2}=\mathrm{d}\mathrm{x}$ 

Integrating both sides, we get 

$\int \frac{\mathrm{d}\mathrm{y}}{2^2+{\mathrm{y}}^2}=\int \mathrm{d}\mathrm{x}$

⇒ $\frac{1}{2}{\tan }^{-1}\frac{\mathrm{y}}{2}=\mathrm{x}+\mathrm{c}$ 

⇒ ${\tan }^{-1}\frac{\mathrm{y}}{2}=2\mathrm{x}+2\mathrm{c}$

⇒ ${\tan }^{-1}\frac{\mathrm{y}}{2}=2\mathrm{x}+\mathrm{C}$  [∵ 2c = C]

⇒ $\frac{\mathrm{y}}{2}=\tan (2\mathrm{x}+\mathrm{C})$

 $\mathrm{y}=2\tan (2\mathrm{x}+\mathrm{C})$ 

The correct option is 2 . 

Concept:

Some useful formulas are:

$\int \frac{\mathrm{d}\mathrm{x}}{\mathrm{a}\mathrm{x}}=\frac{1}{\mathrm{a}}\mathrm{l}\mathrm{o}\mathrm{g}\mathrm{x}+\mathrm{c}$

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

Calculation:

$\frac{\mathrm{d}\mathrm{x}}{\mathrm{d}\mathrm{t}}=3\mathrm{x}+8$

Rearranging the equation and integrating we get,

⇒$\int \frac{\mathrm{d}\mathrm{x}}{3\mathrm{x}+8}=\int \mathrm{d}\mathrm{t}$

⇒$\frac{1}{3}\mathrm{l}\mathrm{o}\mathrm{g}(3\mathrm{x}+8)=\mathrm{t}+\mathrm{c}$, c = constant of integration

⇒ log(3x + 8) = 3(t + c)

⇒ 3x + 8 = e^3(t+c) 

⇒ 3x = e^3(t+c) - 8

∴ $\mathrm{x}=\frac{1}{3}{\mathrm{e}}^{3(\mathrm{t}+\mathrm{c})}-\frac{8}{3}$

Concept:

$\int \frac{\mathrm{d}\mathrm{x}}{\sqrt{{\mathrm{a}}^2-{\mathrm{x}}^2}}=\mathrm{s}\mathrm{i}{\mathrm{n}}^{-1}\frac{\mathrm{x}}{\mathrm{a}}$ 

Calculation:

Given: dy = $\sqrt{1-{\mathrm{y}}^2}$ dx 

⇒ $\frac{\mathrm{d}\mathrm{y}}{\sqrt{1^2-{\mathrm{y}}^2}}=\mathrm{d}\mathrm{x}$ 

Integrating both sides, we get

⇒ $\int \frac{\mathrm{d}\mathrm{y}}{\sqrt{1^2-{\mathrm{y}}^2}}=\int \mathrm{d}\mathrm{x}$ 

⇒ $\mathrm{s}\mathrm{i}{\mathrm{n}}^{-1}\left(\mathrm{y}\right)$ = x + c 

⇒ y = sin ( x + c ) . 

The correct option is 2.

Calculation:

Given: ​$\mathrm{y}\frac{\mathrm{d}\mathrm{y}}{\mathrm{d}\mathrm{x}}=\mathrm{x}+1$

⇒ ydy = (x + 1) dx

Integrating both sides, we get

⇒ ∫ ydy = ∫ (x + 1) dx

⇒ $\frac{{\mathrm{y}}^2}{2}=\frac{{\mathrm{x}}^2}{2}+\mathrm{x}+\mathrm{c}$

⇒ 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:

$\int \frac{1}{{\mathrm{a}}^2+{\mathrm{x}}^2}\mathrm{d}\mathrm{x}=\frac{1}{\mathrm{a}}{\tan }^{-1}\frac{\mathrm{x}}{\mathrm{a}}+\mathrm{c}$

$\int {\mathrm{x}}^{\mathrm{n}}\mathrm{d}\mathrm{x}=\frac{{\mathrm{x}}^{\mathrm{n}+1}}{\mathrm{n}+1}+\mathrm{c}$

Calculation:

Given: $\frac{\mathrm{d}\mathrm{x}}{\mathrm{d}\mathrm{y}}=(1+{\mathrm{x}}^2)(1+{\mathrm{y}}^2)$

$\Rightarrow \frac{\mathrm{d}\mathrm{x}}{(1+{\mathrm{x}}^2)}=(1+{\mathrm{y}}^2)\mathrm{d}\mathrm{y}$

$\Rightarrow (1+{\mathrm{y}}^2)\mathrm{d}\mathrm{y}=\frac{\mathrm{d}\mathrm{x}}{(1+{\mathrm{x}}^2)}$

Integrating both sides, we get

$\Rightarrow \int (1+{\mathrm{y}}^2)\mathrm{d}\mathrm{y}=\int \frac{\mathrm{d}\mathrm{x}}{(1+{\mathrm{x}}^2)}$

$\Rightarrow \mathrm{y}+\frac{{\mathrm{y}}^3}{3}={\tan }^{-1}\mathrm{x}+\mathrm{c}$

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 

$\frac{\mathrm{d}\mathrm{y}}{\mathrm{d}\mathrm{x}}-4\mathrm{y}=0$

⇒ $\frac{\mathrm{d}\mathrm{y}}{\mathrm{y}}=4\mathrm{d}\mathrm{x}$

Integrating both sides

⇒ $\int \frac{\mathrm{d}\mathrm{y}}{\mathrm{y}}=\int 4\mathrm{d}\mathrm{x}$

⇒ ln y = 4x + c

⇒ y = e^{4x + c}

Now y(0) = 1

⇒ 1 = e^{0 + c}

⇒ c = 0

∴ y = e^4x

Concept:

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

$\Rightarrow \int \mathrm{f}\left(\mathrm{x}\right)\mathrm{d}\mathrm{x}=\int \mathrm{g}\left(\mathrm{y}\right)\mathrm{d}\mathrm{y}$

Calculation:

To Find: Solution of the differential equation

$\frac{\mathrm{y}\mathrm{d}\mathrm{x}-\mathrm{x}\mathrm{d}\mathrm{y}}{\mathrm{x}}=0$

⇒ ydx - xdy = 0

⇒ ydx = xdy 

⇒ $\frac{\mathrm{d}\mathrm{y}}{\mathrm{y}}=\frac{\mathrm{d}\mathrm{x}}{\mathrm{x}}$

Integrating both sides, we get

$$\begin{gathered} \int \frac{\mathrm{d}\mathrm{y}}{\mathrm{y}}=\int \frac{\mathrm{d}\mathrm{x}}{\mathrm{x}} \\ \Rightarrow \ln \mathrm{y}=\ln \mathrm{x}+\ln \mathrm{c} \end{gathered}$$   

$\Rightarrow \ln (\mathrm{y})=\ln \mathrm{c}\mathrm{x}$                   (∵ ln x + ln y = ln (xy))

⇒ y = cx 

∴ y - cx = 0