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Differential Equations — Maths STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 ScienceMathsDifferential Equations1 Mark
MCQ
The solution of the differential equartion $y_1y_3 = y_2$ is:
  • A
    $\text{x}=\text{C}_{1}\text{e}^{\text{C}_{2}\text{y}}+\text{C}_{3}$
  • ✓
    $\text{y}=\text{C}_{1}\text{e}^{\text{C}_{2}\text{x}}+\text{C}_{3}$
  • C
    $2\text{x}=\text{C}_{1}\text{e}^{\text{C}_{2}\text{y}}+\text{C}_{3}$
  • D
    None of these.
✓

Answer

Correct option: B.
$\text{y}=\text{C}_{1}\text{e}^{\text{C}_{2}\text{x}}+\text{C}_{3}$
$\text{y}_{1}\text{y}_{3}=\text{y}^{2}_{2}$
$\frac{\text{y}_{3}}{\text{y}_{2}}=\frac{\text{y}_{2}}{\text{y}_{1}}$
$\Rightarrow \frac{\Big(\frac{\text{d}^{3}\text{y}}{\text{dx}^{3}}\Big)}{\Big(\frac{\text{d}^{2}\text{y}}{\text{dx}^{2}}\Big)}=\frac{\Big(\frac{\text{d}^{2}\text{y}}{\text{dx}^{2}}\Big)}{\Big(\frac{\text{d}\text{y}}{\text{dx}}\Big)}$
$\Rightarrow\int \frac{\frac{\text{d}}{\text{dx}}\Big(\frac{\text{d}^{3}\text{y}}{\text{dx}^{3}}\Big)}{\Big(\frac{\text{d}^{2}\text{y}}{\text{dx}^{2}}\Big)}=\int\frac{\frac{\text{d}}{\text{dx}}\Big(\frac{\text{d}\text{y}}{\text{dx}}\Big)}{\Big(\frac{\text{d}\text{y}}{\text{dx}}\Big)}$
$\Rightarrow\log\Big(\frac{\text{d}^{2}\text{y}}{\text{dx}^{2}}\Big)=\log\Big(\frac{\text{dy}}{\text{dx}}\Big)+\log\text{C}_{4}$
$\Rightarrow \frac{\text{d}^{2}\text{y}}{\text{dx}}^{2}=\text{C}_{4}\frac{\text{dy}}{\text{dx}}$
$\Rightarrow\int \frac{\frac{\text{d}}{\text{dx}}\Big(\frac{\text{d}^{3}\text{y}}{\text{dx}^{3}}\Big)}{\Big(\frac{\text{d}^{2}\text{y}}{\text{dx}^{2}}\Big)}=\int\text{C}_{4}\ \text{dx}$
$\Rightarrow \log\Big(\frac{\text{dy}}{\text{dx}}\Big)=\text{C}_{4}\text{x}+\text{C}_{5}$
$\Rightarrow \int\text{dy}=\int\text{e}^{\text{C}_{4}\text{x+}\text{C}_{5}}\ \text{dx}$
$\Rightarrow\text{y}=​​\frac{\text{e}^{\text{C}_{4}\text{x}+\text{C}_{3}}}{\text{C}_{4}}+\text{C}_{6}$
$\Rightarrow\text{y}=​\text{C}_{1}\text{e}^{\text{C}_{2}\text{x}}+\text{C}_{3}$
Where,
$\text{C}_{1}=\frac{\text{e}^{\text{C}_{5}}}{\text{C}_{4}}$
$\text{C}_{4}=\text{C}_{2}$
$\text{C}_{6}=\text{C}_{3}$

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