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

Hint : First we have to find the first and second order derivatives of the given function. Then consider the function $\dfrac{{{d^2}y}}{{d{x^2}}} - 5\dfrac{{dy}}{{dx}} + 6y$ .Substitute the values of the function $y$ and their derivatives in the consider function and prove that is equal to zero.

Complete step-by-step answer :
A differential equation is the equation which contains dependent variables, independent variables and derivatives of the dependent variables with respect to the independent variables.
Suppose $u$ and $v$ are two functions. Then differentiation has the following rules.
1. $\dfrac{d}{{dx}}\left( {u \times v} \right) = v \times \dfrac{d}{{dx}}\left( u \right) + u \times \dfrac{d}{{dx}}\left( v \right)$ .
2. $\dfrac{d}{{dx}}\left( {\dfrac{u}{v}} \right) = \dfrac{{v \times \dfrac{d}{{dx}}\left( u \right) - u \times \dfrac{d}{{dx}}\left( v \right)}}{{{v^2}}}$ .
Given $y = 3{e^{2x}} + 2{e^{3x}}$ ----(1)
Differentiating with respect to $x$ both sides of the equation (1), we get
$\dfrac{{dy}}{{dx}} = 6{e^{2x}} + 6{e^{3x}}$ -------(2)
Differentiating with respect to $x$ both sides of the equation (2), we get
$\dfrac{{{d^2}y}}{{d{x^2}}} = 12{e^{2x}} + 18{e^{3x}}$ ------(3)
Consider $\dfrac{{{d^2}y}}{{d{x^2}}} - 5\dfrac{{dy}}{{dx}} + 6y$
Substitute the values of $y$ , $\dfrac{{dy}}{{dx}}$ , and $\dfrac{{{d^2}y}}{{d{x^2}}}$ from the equations (1),(2), and (3) in the above consider expression, we get
$\dfrac{{{d^2}y}}{{d{x^2}}} - 5\dfrac{{dy}}{{dx}} + 6y = 12{e^{2x}} + 18{e^{3x}} - 5\left( {6{e^{2x}} + 6{e^{3x}}} \right) + 6\left( {3{e^{2x}} + 2{e^{3x}}} \right)$
Simplifying the RHS of the above equation, we get
$\dfrac{{{d^2}y}}{{d{x^2}}} - 5\dfrac{{dy}}{{dx}} + 6y = 12{e^{2x}} + 18{e^{3x}} - 30{e^{2x}} + 30{e^{3x}} + 18{e^{2x}} + 12{e^{3x}} = 0$
Hence, we prove $\dfrac{{{d^2}y}}{{d{x^2}}} - 5\dfrac{{dy}}{{dx}} + 6y = 0$ when $y = 3{e^{2x}} + 2{e^{3x}}$ .

Note : Note that the derivative of the constant times the function is equal to constant times the derivative of the given function. Since differential equations are classified into two types, Ordinary differential equations where dependent variables depend on only one independent variable and Partial differential equations where dependent variables depend on two or more independent variables.