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Differential Equations Ncert Solutions Maths class 12th Maths Class 12th Differential Equations

27. $y = a e^{3 x} + b e^{- 2 x}$

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Vishal Baghel | Contributor-Level 10

4 months ago

Given: Equation of the family of curves $y = a e^{3 x} + b e^{- 2 x} . . . . . . . . . . (i)$

Differentiating both sides with respect to x, we get:

$y^{'} = 3 a e^{3 x} - 2 b e^{- 2 x} . . . . . . . . . . (i i)$

Again, differentiating both sides with respect to x, we get:

$y^{"} = 9 a e^{3 x} - 4 b e^{- 2 x} . . . . . . . . . . (i i i)$

Multiplying equation (i) with (ii) and then adding it to equation (ii), we get:

$\begin{array}{l} (2 a e^{3 x} + 2 b e^{- 2 x}) + (3 a e^{3 x} - 2 b e^{- 2 x}) = 2 y + y^{'} \\ \Rightarrow 5 a e^{3 x} = 2 y + y^{'} \\ \Rightarrow a e^{3 x} = \frac{2 y + y^{'}}{5} \end{array}$

Now, multiplying equation (i) with (iii) and subtracting equation (ii) from it, we get:

$\begin{array}{l} (3 a e^{3 x} + 2 b e^{- 2 x}) - (3 a e^{3 x} - 2 b e^{- 2 x}) = 3 y - y^{'} \\ \Rightarrow 5 b e^{- 2 x} = 3 y - y^{'} \\ \Rightarrow b e^{- 2 x} = \frac{3 y - y^{'}}{5} \end{array}$

Substituting the values of $a e^{3 x}$ and $b e^{- 2 x}$ in equation (iii), we get:

$\begin{array}{l} y^{"} = 9 . \frac{(2 y - y^{'})}{5} + 4 \frac{(3 y - y^{'})}{5} \\ \Rightarrow y^{"} = \frac{18 y + 9 y^{'}}{5} + \frac{12 y - 4 y^{'}}{5} \\ \Rightarrow y^{"} = \frac{30 y + 5 y^{'}}{5} \\ \Rightarrow y^{"} = 6 y + y^{'} \\ \Rightarrow y^{"} - y^{'} - 6 y = 0 \end{array}$

This is the required differential equation of the given curve.

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