# Show that

Question:

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

Solution:

$y=a e^{3 x}+b e^{-2 x}$                  ...(1)

Differentiating both sides with respect to x, we get:

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

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

$y^{\prime \prime}=9 a e^{3 x}+4 b e^{-2 x}$          ...(3)

Multiplying equation (1) with (2) and then adding it to equation (2), we get:

$\left(2 a e^{3 x}+2 b e^{-2 x}\right)+\left(3 a e^{3 x}-2 b c^{-2 x}\right)=2 y+y^{\prime}$

$\Rightarrow 5 a e^{3 x}=2 y+y^{\prime}$

$\Rightarrow a e^{3 x}=\frac{2 y+y^{\prime}}{5}$

Now, multiplying equation (1) with 3 and subtracting equation (2) from it, we get:

$\left(3 a e^{3 x}+3 b e^{-2 x}\right)-\left(3 a e^{3 x}-2 b e^{-2 x}\right)=3 y-y^{\prime}$

$\Rightarrow 5 b e^{-2 x}=3 y-y^{\prime}$

$\Rightarrow b e^{-2 x}=\frac{3 y-y^{\prime}}{5}$

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

$y^{\prime \prime}=9 \cdot \frac{\left(2 y+y^{\prime}\right)}{5}+4 \frac{\left(3 y-y^{\prime}\right)}{5}$

$\Rightarrow y^{\prime \prime}=\frac{18 y+9 y^{\prime}}{5}+\frac{12 y-4 y^{\prime}}{5}$

$\Rightarrow y^{\prime \prime}=\frac{30 y+5 y^{\prime}}{5}$

$\Rightarrow y^{\prime \prime}=6 y+y^{\prime}$

$\Rightarrow y^{\prime \prime}-y^{\prime}-6 y=0$

This is the required differential equation of the given curve.