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Differential Equations — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsDifferential Equations1 Mark
MCQ
The differential equation of the ellipse $\frac{\text{x}^{2}}{\text{a}^{2}}+\frac{\text{y}^{2}}{\text{b}^{2}}=\text{C}$ is:
  • ✓
    $\frac{\text{y}''}{\text{y}'}+\frac{\text{y}'}{\text{y}}-\frac{1}{\text{x}}=0$
  • B
    $\frac{\text{y}''}{\text{y}'}+\frac{\text{y}'}{\text{y}}+\frac{1}{\text{x}}=0$
  • C
    $\frac{\text{y}''}{\text{y}'}-\frac{\text{y}'}{\text{y}}-\frac{1}{\text{x}}=0$
  • D
    None of these.
✓

Answer

Correct option: A.
$\frac{\text{y}''}{\text{y}'}+\frac{\text{y}'}{\text{y}}-\frac{1}{\text{x}}=0$
We have,
$\frac{\text{x}^{2}}{\text{a}^{2}}+\frac{\text{y}^{2}}{\text{b}^{2}}=\text{C}\ ...(\text{i})$
Differentiating with respect to x, we get
$\frac{2\text{x}^{2}}{\text{a}^{2}}+\frac{2\text{y}^{2}}{\text{b}^{2}}\text{y}'=0$
$\frac{\text{x}^{2}}{\text{a}^{2}}+\frac{\text{y}^{2}}{\text{b}^{2}}\text{y}'=0\ ...(\text{ii})$
Again differentiating with respect to x, we get
$\Rightarrow \frac{1}{\text{a}^{2}}+\frac{1}{\text{b}^{2}}(\text{y'}^{2})+\frac{\text{xy}}{\text{b}^{2}}\text{y}''=0\ ...(\text{iii})$
Multiplying throughout by x, we get
$\Rightarrow \frac{\text{x}}{\text{a}^{2}}+\frac{\text{x}}{\text{b}^{2}}(\text{y'}^{2})+\frac{\text{xy}}{\text{b}^{2}}\text{y}''=0\ ...(\text{iv})$
Subtracting (ii) from (iv),
$\frac{1}{\text{b}^{2}}\big[\text{x}(\text{y}')^{2}+\text{xyy}''-\text{yy}''\big]=0$
$\Rightarrow \text{x}(\text{y}')^{2}+\text{xyy}''-\text{yy}''=0$
Diving both sides by,
$\frac{\text{y}''}{\text{y}'}+\frac{\text{y}'}{\text{y}}-\frac{1}{\text{x}}=0$
$\Rightarrow \frac{\text{y}''}{\text{y}'}+\frac{\text{y}'}{\text{y}}-\frac{1}{\text{x}}=0$

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