Question

If \[y = P{e^{ax}} + Q{e^{bx}}\], then show that \[\dfrac{{{d^2}y}}{{d{x^2}}} - (a + b)\dfrac{{dy}}{{dx}} + aby = 0\].

Answer 1

Hint: We have the value of y, find the value of \[\dfrac{{dy}}{{dx}}\] and \[\dfrac{{{d^2}y}}{{d{x^2}}}\] , and then substitute the values in the given expression to show that \[\dfrac{{{d^2}y}}{{d{x^2}}} - (a + b)\dfrac{{dy}}{{dx}} + aby = 0\].

Complete step-by-step answer:
A differential equation is an equation that relates one or more functions and their derivatives.
Any function that satisfies the differential equation is said to be the solution of the differential equation.
A differential equation can have more than one solution and the general form of the solution is called the general solution of the differential equation.

We are given the value of y as follows:
\[y = P{e^{ax}} + Q{e^{bx}}............(1)\]
And we need to show the following is true:
\[\dfrac{{{d^2}y}}{{d{x^2}}} - (a + b)\dfrac{{dy}}{{dx}} + aby = 0............(2)\]
For this to be true, \[y = P{e^{ax}} + Q{e^{bx}}\] must be a solution of \[\dfrac{{{d^2}y}}{{d{x^2}}} - (a + b)\dfrac{{dy}}{{dx}} + aby = 0\].
We can find the values of \[\dfrac{{dy}}{{dx}}\] and \[\dfrac{{{d^2}y}}{{d{x^2}}}\] and substitute in the expression \[\dfrac{{{d^2}y}}{{d{x^2}}} - (a + b)\dfrac{{dy}}{{dx}} + aby\] and check if it is equal to zero.
Differentiating both sides of equation (1) with respect to x, we have:
\[\dfrac{{dy}}{{dx}} = \dfrac{d}{{dx}}(P{e^{ax}} + Q{e^{bx}})\]
We know that the derivative of the sum of two functions is the sum of the derivative of the two functions. Hence, we have:
\[\dfrac{{dy}}{{dx}} = \dfrac{d}{{dx}}(P{e^{ax}}) + \dfrac{d}{{dx}}(Q{e^{bx}})\]
The derivative of \[{e^{ax}}\] is \[a{e^{ax}}\], then, we have:
\[\dfrac{{dy}}{{dx}} = aP{e^{ax}} + bQ{e^{bx}}..........(3)\]
Differentiating both sides of equation (3) with respect to x, we have:
\[\dfrac{{{d^2}y}}{{d{x^2}}} = \dfrac{d}{{dx}}(aP{e^{ax}} + bQ{e^{bx}})\]
Simplifying, we have:
\[\dfrac{{{d^2}y}}{{d{x^2}}} = \dfrac{d}{{dx}}(aP{e^{ax}}) + \dfrac{d}{{dx}}(bQ{e^{bx}})\]
\[\dfrac{{{d^2}y}}{{d{x^2}}} = {a^2}P{e^{ax}} + {b^2}Q{e^{bx}}..........(4)\]
Let \[LHS = \dfrac{{{d^2}y}}{{d{x^2}}} - (a + b)\dfrac{{dy}}{{dx}} + aby\], then substituting equations (1), (3) and (4) in equation (2), we have:
\[LHS = {a^2}P{e^{ax}} + {b^2}Q{e^{bx}} - (a + b)(aP{e^{ax}} + bQ{e^{bx}}) + ab(P{e^{ax}} + Q{e^{bx}})\]
Evaluating, we have:
\[LHS = {a^2}P{e^{ax}} + {b^2}Q{e^{bx}} - {a^2}P{e^{ax}} - abP{e^{ax}} - {b^2}Q{e^{bx}} - abQ{e^{bx}} + abP{e^{ax}} + abQ{e^{bx}}\]
Canceling common terms, we have:
\[LHS = 0\]
\[\dfrac{{{d^2}y}}{{d{x^2}}} - (a + b)\dfrac{{dy}}{{dx}} + aby = 0\]
Hence, we showed \[\dfrac{{{d^2}y}}{{d{x^2}}} - (a + b)\dfrac{{dy}}{{dx}} + aby = 0\].

Note: Do not forget the negative sign when evaluating \[ - (a + b)\dfrac{{dy}}{{dx}}\] and remember that the differentiation of \[a{e^{ax}}\] is \[{a^2}{e^{ax}}\].Students should remember the formulas of product of differentiation and quotient of differentiation for solving these types of problems.