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DIFFERENTIAL EQUATIONS — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsDIFFERENTIAL EQUATIONS4 Marks
Question
A differential equation is said to be in the variable separable form if it is expressible in the form f(x) dx = g(y) dy.
The solution of this equation is given by
$\int\text{f(x)dx}=\int\text{g(y)dy}+\text{c},$ where c is the constant of integration.
Based on the above information, answer the following questions.
  1. If the solution of the differential equation $\frac{\text{dy}}{\text{dx}}=\frac{\text{ax+3}}{\text{2y+f}}$ represents a circle, then the value of 'a' is:
  1. 2
  2. -2
  3. 3
  4. -4
  1. The differential equation $\frac{\text{dy}}{\text{dx}}=\frac{\sqrt{1-\text{y}^2}}{\text{y}}$ determines a family of circle with.
  1. Variable radii and fixed centre (0, 1)
  2. Variable radii and fixed centre (0, -1)
  3. Fixed radius 1 and variable centre on x-axis
  4. Fixed radius 1 and variable centre on y-axis
  1. If = y'+ 1, y(0) = 1, then y (In 2) =
  1. 1
  2. 2
  3. 3
  4. 4
  1. The solution of the differential equation $\frac{\text{dy}}{\text{dx}}=\text{e}^\text{x-y}+\text{x}^2\text{e}^\text{-y}$ is:
  1. $\text{e}^\text{x}=\frac{\text{y}^3}{3}+\text{e}^\text{y}+\text{c}$
  2. $\text{e}^\text{y}=\frac{\text{x}^2}{3}+\text{e}^\text{x}+\text{c}$
  3. $\text{e}^\text{y}=\frac{\text{x}^3}{3}+\text{e}^\text{x}+\text{c}$
  4. None of these
  1. If $\frac{\text{dy}}{\text{dx}}=\text{y}\sin2\text{x},\ \text{y}(0)=1,$ then its solution is:
  1. $\text{y}=\text{e}^{\sin^2}\text{x}$
  2. $\text{y}={\sin^2}\text{x}$
  3. $\text{y}={\cos^2}\text{x}$
  4. $\text{y}=\text{e}^{\cos^2}\text{x}$
✓

Answer

  1. (b) -2

Solution:

We have, $\frac{\text{dy}}{\text{dx}}=\frac{\text{ax+3}}{\text{2y+f}}$

$\Rightarrow\ \ (\text{ax+3})\text{dx}=(2\text{y}+\text{f})\text{dy}$

$\Rightarrow\text{a}\frac{\text{x}^2}{2}+\text{3x}=\text{y}^2+\text{fy}+\text{c}$ (Integrating)

$\Rightarrow-\frac{\text{a}}{2}\text{x}^2+\text{y}^2-\text{3x}+\text{fy}+\text{C}=0$

This will represent a circle, if $\frac{-\text{a}}{2}=1\Rightarrow\text{a}=-2$

[ $\therefore$ In circle, coefficient of x2 = coefficient of y2)

  1. (c) Fixed radius 1 and variable centre on x-axis

Solution:

We have, $\frac{\text{ydy}}{\sqrt{1-\text{y}^2}}=\text{dx}$

On integration, we get $-\sqrt{1-\text{y}^2}=\text{x+c}$

⇒ 1 - y2 = (x + c)2 ⇒ (x + c)2 + y2 = 1 which represents a circle with radius I and centre on the x-axis.

  1. (c) 3

Solution:

$\text{y}'=\text{y}+1\Rightarrow\frac{\text{dy}}{\text{y}+1}=\text{dx}$

⇒ In (y + 1) = x + c (integrating)

Now, y(0) = 1 ⇒ c = In 2

$\therefore \ \text{In}\Bigg(\frac{\text{y}+1}{2}\Bigg)=\text{x}$

⇒ y + 1 = 2ex

So, y (In 2) = -1 + 2eIn 2 = -1 + 4 = 3

  1. (c) $\text{e}^\text{y}=\frac{\text{x}^3}{3}+\text{e}^\text{x}+\text{c}$

Solution:

From the given differential equation, we have

$\frac{\text{dy}}{\text{dx}}=\frac{\text{e}^\text{x}+\text{x}^\text{2}}{\text{e}^\text{y}}$

$\Rightarrow\ \text{e}^\text{y}\text{dy}=(\text{e}^\text{x}+\text{x}^2)\text{dx}$

Integrating, we get $\text{e}^\text{y}=\text{e}^\text{x}+\frac{\text{x}^3}{3}+\text{c}$

  1. (a) $\text{y}=\text{e}^{\sin^2}\text{x}$

Solution:

We have, $\frac{\text{dy}}{\text{dx}}=\text{y}\sin2\text{x}$

$\Rightarrow\ \frac{\text{dy}}{\text{y}}=\sin2\text{x}\ \text{dx}$

$\Rightarrow\ \log\text{y}=-\frac{\cos2\text{x}}{2}+\text{c}$

Since x = 0, y = 1

therefore $\text{C}=\frac{1}{2}$

Now, $\log\text{y}=\frac{1}{2}(1-\cos2\text{x})$

$\Rightarrow\ \log\text{y}=\sin^2\text{x}\Rightarrow\text{y}=\text{e}^{\sin^2}\text{x}$

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