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Differential Equations — MATHS STD 12 Science — Question

Rajasthan 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 $x^2 =$ coefficient of $y^2)$
  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 $-\sqrt{1-\text{y}^2}=\text{x+c}$
$\Rightarrow 1 - y^2 = (x + c)^{2}​​​​​​​$
$\Rightarrow ^(x + c)^{2 }+ y^{2 }= 1$ which represents a circle with radius $I$ and centre on the $x-$axis.
  1. $(c)\ 3$
Solution:
$\text{y}'=\text{y}+1\Rightarrow\frac{\text{dy}}{\text{y}+1}=\text{dx}$
$\Rightarrow $ In $(y + 1) = x + c ($integrating$)$
Now, $y(0) = 1 \Rightarrow c = In\ 2$
$\therefore \ \text{In}\Bigg(\frac{\text{y}+1}{2}\Bigg)=\text{x}$
$\Rightarrow y + 1 = 2e^x$​​​​​​​
So, $y (In\ 2) = -1 + 2e^{In\ 2} = -1 + 4 = 3$
  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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