Download App

DIFFERENTIAL EQUATIONS — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsDIFFERENTIAL EQUATIONS1 Mark
MCQ
The differention equation $\frac{\text{dy}}{\text{dx}}+\text{P}\text{y}=\text{Qy}^{\text{n}},\text{n}>2$ can be reduced to linear from by substituting:
  • A
    $\text{z}=\text{y}^{\text{n}-1}$
  • B
    $\text{z}=\text{y}^{\text{n}}$
  • C
    $\text{z}=\text{y}^{\text{n}+1}$
  • ✓
    $\text{z}=\text{y}^{1-\text{n}}$
✓

Answer

Correct option: D.
$\text{z}=\text{y}^{1-\text{n}}$
We have,
$\frac{\text{dy}}{\text{dx}}+\text{P}\text{y}=\text{Qy}^{\text{n}}$
$\Rightarrow \text{y}^{-\text{n}}\frac{\text{dy}}{\text{dx}}+\text{Py}^{1-\text{n}}=\text{Q}\ ...(\text{i})$
Put $\text{z}=\text{y}^{1-\text{n}}$
Integrating both sides with respect to x, we get
$\frac{\text{dz}}{\text{dx}}=(1-\text{n})\text{y}^{-\text{n}}\frac{\text{dy}}{\text{dx}}$
$\Rightarrow \text{y}^{-\text{n}}\frac{\text{dy}}{\text{dx}}=\frac{1}{(1-\text{n})}\frac{\text{dz}}{\text{dx}}$
Now, (i),
$\frac{1}{(1-\text{n})}\frac{\text{dz}}{\text{dx}}+\text{Pz}=\text{Q}$
$\Rightarrow \frac{\text{dz}}{\text{dx}}+\text{P}(1-\text{n})=\text{Q}(1-\text{n})$
Which is linear from of differential equation.
Therefore the given differential equation can be to linear by the $\text{z}=\text{y}^{1-\text{n}}.$

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free

Explore more

More "M.C.Q (1 Marks)" from DIFFERENTIAL EQUATIONS→DIFFERENTIAL EQUATIONS — all question types→Maths STD 12 Science question papers→Gujarat Board STD 12 Science question papers→Gujarat Board question paper generator→

Similar questions

Let $f$ : $R \rightarrow R$ be a differentiable function $ \&$ $ f (1) = 4$  , then the value of $\mathop {Lim}\limits_{x\,\, \to \,\,1} $$\int\limits_4^{f\,(x)} {\,\frac{{2t\,dt}}{{x - 1}}} $; is :
→
Direction ratios of a vector parallel to line $\frac{x-1}{2}=-y=\frac{2 z+1}{6}$ are :
→
The area of a triangle is $5$ and two of its vertices are $A(2, 1), B(3, -2)$. The third  vertex which lies on line $y = x + 3$ is-
→
Side of an equilateral triangle expands at the rate of $2\, cm/s.$ The rate of increase of its area when each side is $10\, cm,$ is
→
Which of the following is not correct in a given determinant of $A,$ where $A = [a_{ij}]_{3\times 3}$:
→
If $y = {{{{(1 - x)}^2}} \over {{x^2}}}$, then ${{dy} \over {dx}}$ is
→
Which of the following is true?
→
Consider the function $f (x) =$$\left[ \begin{gathered}   \hfill \\   \hfill \\   \hfill \\   \hfill \\   \hfill \\  \end{gathered}  \right.$$\begin{gathered}  \frac{x}{{[x]}}\;\;\;\;\;\;\;\;\;\;\;\;\;if\;\;1 \leqslant \;x < 2 \hfill \\   \hfill \\  1\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;if\;\;x = 2 \hfill \\   \hfill \\  \sqrt {6 - x} \;\;\;\;\;\;\;if\;\;2 < x \leqslant 3 \hfill \\ \end{gathered} $ 

where $[x]$ denotes step up function then at $x = 2$ function

→
Evaluate $\begin{bmatrix}1&0&1\\0&0&1\\1&0&1\end{bmatrix}$ is:
→
The domain of the function

$f(x)=\sin ^{-1}\left(\frac{3 x^{2}+x-1}{(x-1)^{2}}\right)+\cos ^{-1}\left(\frac{x-1}{x+1}\right)$ is :

→