Question

If $y = 3{e^{2x}} + 2{e^{3x}}$. Prove that $\dfrac{{{d^2}y}}{{d{x^2}}} - 5\dfrac{{dy}}{{dx}} + 6y = 0$

Answer 1

Here, we will prove the given condition. We will find the first derivative of the function, then the second derivative of the function. By substituting the first derivative of the function, the second derivative of the function and the function in the given condition, we will prove the given condition.

Formula Used:
We will use the following derivative formula:
1.$\dfrac{d}{{dx}}\left( {{e^x}} \right) = {e^x}$
2.$\dfrac{d}{{dx}}\left( x \right) = x$

Complete step-by-step answer:
We are given that $y = 3{e^{2x}} + 2{e^{3x}}$.
$\Rightarrow y = 3{e^{2x}} + 2{e^{3x}}$………………………………………………………….$\left( 1 \right)$
Now, we will differentiate equation $\left( 1 \right)$with respect to $x$ .
Differentiating equation $\left( 1 \right)$with respect to $x$ by using the derivative formula, we get
$\Rightarrow \dfrac{{dy}}{{dx}} = 3{e^{2x}} \cdot 2\left( 1 \right) + 2{e^{3x}} \cdot 3\left( 1 \right)$
By multiplying the terms, we get
$\Rightarrow \dfrac{{dy}}{{dx}} = 6{e^{2x}} + 6{e^{3x}}$
By taking the common factor, we get
$\Rightarrow \dfrac{{dy}}{{dx}} = 6\left( {{e^{2x}} + {e^{3x}}} \right)$………………………………………………………$\left( 2 \right)$
Now, we will differentiate equation $\left( 2 \right)$with respect to $x$ .
Differentiating equation $\left( 2 \right)$with respect to $x$ by using the derivative formula, we get
$\Rightarrow \dfrac{{{d^2}y}}{{d{x^2}}} = 6\left( {{e^{2x}} \cdot 2\left( 1 \right) + {e^{3x}} \cdot 3\left( 1 \right)} \right)$

By multiplying the terms, we get
$\Rightarrow \dfrac{{{d^2}y}}{{d{x^2}}} = 6\left( {2{e^{2x}} + 3{e^{3x}}} \right)$………………………………………………………………………………$\left( 3 \right)$
Now, we will prove that $\dfrac{{{d^2}y}}{{d{x^2}}} - 5\dfrac{{dy}}{{dx}} + 6y = 0$
Now, by substituting the first derivative, the second derivative and the given function, we get
$\Rightarrow \dfrac{{{d^2}y}}{{d{x^2}}} - 5\dfrac{{dy}}{{dx}} + 6y = 0$
$\Rightarrow \dfrac{{{d^2}y}}{{d{x^2}}} - 5\dfrac{{dy}}{{dx}} + 6y = 6\left( {2{e^{2x}} + 3{e^{3x}}} \right) - 5\left( {6\left( {{e^{2x}} + {e^{3x}}} \right)} \right) + 6\left( {3{e^{2x}} + 2{e^{3x}}} \right)$
By multiplying the terms, we get
$\Rightarrow \dfrac{{{d^2}y}}{{d{x^2}}} - 5\dfrac{{dy}}{{dx}} + 6y = 6\left( {2{e^{2x}} + 3{e^{3x}}} \right) - 5\left( {6{e^{2x}} + 6{e^{3x}}} \right) + 6\left( {3{e^{2x}} + 2{e^{3x}}} \right)$
By multiplying the terms, we get
$\Rightarrow \dfrac{{{d^2}y}}{{d{x^2}}} - 5\dfrac{{dy}}{{dx}} + 6y = 12{e^{2x}} + 18{e^{3x}} - 30{e^{2x}} - 30{e^{3x}} + 18{e^{2x}} + 12{e^{3x}}$
By adding the terms, we get
$\Rightarrow \dfrac{{{d^2}y}}{{d{x^2}}} - 5\dfrac{{dy}}{{dx}} + 6y = 30{e^{2x}} + 30{e^{3x}} - 30{e^{2x}} - 30{e^{3x}}$
By subtracting the terms, we get
$\Rightarrow \dfrac{{{d^2}y}}{{d{x^2}}} - 5\dfrac{{dy}}{{dx}} + 6y = 0$
We have proved that $\dfrac{{{d^2}y}}{{d{x^2}}} - 5\dfrac{{dy}}{{dx}} + 6y = 0$.
Therefore, if $y = 3{e^{2x}} + 2{e^{3x}}$, then $\dfrac{{{d^2}y}}{{d{x^2}}} - 5\dfrac{{dy}}{{dx}} + 6y = 0$.

Note: We know that differentiation is defined as a rate of change in function based on one of its variables only. We should remember some rules in differentiation which includes that the derivative of a constant is always zero. The derivative of a combination of functions should be followed in an order that the differentiation of the exponential function at first followed by the differentiation of the function of the algebraic function. The derivative of a constant multiplied by a function is equal to the constant multiplied by the derivative of the function.