Question

How to solve the initial value problem?
$y'' + 2y' + y = 0,y(0) = 4,y'(0) = - 6$

Answer 1

Hint : Differential equation: It is an equation that relates one or more functions and their derivative.
Second order differential equation: It is an ordinary differential equation that may be written in the form. $x''(t) = F(t,x(t),x'(t))$ for some function $F$ of three variables.
For a homogeneous differential equation.
The general solution is given by $y(t) = {c_1}{e^{{m_1}x}} + {c_2}{e^{{m_2}x}}$
The general solution having common roots is given by $y(x) = {c_1}{e^{mx}} + {c_2}x{e^{mx}}$ .

Complete step by step solution:
As we know, go through the question.
As it is a homogeneous differential equation.
As it $f(0).y = 0$
The characteristic equation of this ODE is ${D^2}y + 2Dy + y = 0$ .
To get the solution of the ODE. We have to solve the quadratic equation:
${D^2}y + 2Dy + y = 0$
So, we can write auxiliary equation as,
$f(m) = 0$
${m^2}y + 2my + y = 0$
(By using ${(a + b)^2} = {a^2} + 2ab + {b^2}$ )
So, we can write auxiliary equation as,
$\Rightarrow {\left( {m + 1} \right)^2}y = 0$
So, the root of the equation is the same and it is $m = - 1$ .
So, keeping it in general solution format. We get,
$y(x) = {c_1}{e^{mx}} + {c_2}x{e^{mx}}$
$\Rightarrow y(x) = {c_1}{e^{ - 1.x}} + {c_2}x{e^{ - 1.x}} = {e^{ - 1.x}}({c_1} + x{c_2})$
As condition given $y(0) = 4,y'(0) = - 6$
Keeping it step wise we get,
$\Rightarrow y(x) = {e^{ - x}}({c_1} + x.{c_2})$
Keeping $y(0) = 4$ . We get,
$\Rightarrow y(0) = 4 = {e^0}({c_1} + 0.{c_2})$
$\Rightarrow 4 = {c_1}$
Differentiate it w.r.t to $x$ .
$\Rightarrow y(x) = - {c_1}{e^{ - x}} - {c_2}x{e^{ - x}} + {c_2}{e^{ - x}}$
Keeping $y'(0) = - 6$ ,
$\Rightarrow y(0) = - 6 = - {c_1}{e^0} - {c_2}0{e^0} + {c_2}{e^0}$
$\Rightarrow - 6 = - {c_1} + {c_2}$
Keeping values from above calculation.
$\Rightarrow - 6 = - 4 + {c_2}$
$\Rightarrow {c_2} = -2$
So, keeping it in solution. We get,
$\Rightarrow y(x) = 4{e^{mx}} - 2x{e^{mx}}$
Hence the solution of the given differential equation will be $y(x) = 4{e^{mx}} + 2x{e^{mx}}$
So, the correct answer is “ $\ y(x) = 4{e^{mx}} - 2x{e^{mx}}$ ”.

Note : The study of differential equations is a wide field in pure and applied mathematics, physics, and engineering. All of these disciplines are concerned with the properties of equations of various types. Pure mathematics focuses on the existence and uniqueness of solutions, while applied mathematics emphasizes the rigorous justification of the methods for approximating solutions.