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Differential Equations — Maths STD 12 Science — Question

Maharashtra 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 $\text{z}=\text{y}^{1-\text{n}}.$

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