-b.3.1.9 Find the solution of y''+y'-2y=0\quad y(0)=1; y'(0)=1.

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In summary, the differential equation $y''+y'-2y=0$ with initial conditions $y(0)=1$ and $y'(0)=1$ has the solution $y=e^t$.
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karush
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$\textit{given $y''+y'-2y=0\quad y(0)=1 \quad y'(0)=1$ }$
\begin{align*}\displaystyle
\textit{if } r&=e^{t} \textit{then:}\\
y''+y'-2y&=r^2+r-2=(r+2)(r-1)=0\\
&=c_1^{2t}+c_2^{-t}\\
y&=\color{red} {e^t; \quad y \to \infty \,as \, t \to \infty }
\end{align*}ok the bk answer is in red
I just don't get this
why do we need $$y(0)=1 \quad y'(0)=1$$

the graph of $y=e^t$

3.1.9.png
 
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  • #2
karush said:
$\textit{given $y''+y'-2y=0\quad y(0)=1 \quad y'(0)=1$ }$
\begin{align*}\displaystyle
\textit{if } r&=e^{t} \textit{then:}\\
y''+y'-2y&=r^2+r-2=(r+2)(r-1)=0\\
&=c_1^{2t}+c_2^{-t}\\
y&=\color{red} {e^t; \quad y \to \infty \,as \, t \to \infty }
\end{align*}ok the bk answer is in red
I just don't get this
why do we need $$y(0)=1 \quad y'(0)=1$$
The auxiliary equation $r^2 + r - 2 = 0$ factorises as $(r+2)(r-1)=0$, with solutions $r = -2$ and $r=1$. That tells you that the differential equation has solutions $y = e^{-2t}$ and $y = e^t$. The general solution is then $y = c_1e^{-2t} + c_2e^t$, where $c_1$ and $c_2$ are constants. To find the value of those constants, you need to use the conditions $y(0)=1$ and $y'(0)=1$. In fact, if you put $t=0$ in the equation $y = c_1e^{-2t} + c_2e^t$, it tells you that $1 = c_1 + c_2$. If you then differentiate the formula for $y$ and again put $t=0$, it will give you another equation for $c_1$ and $c_2$. You should find that $c_1=0$ and $c_2=1$, so that the solution becomes $y=e^t$.
 

FAQ: -b.3.1.9 Find the solution of y''+y'-2y=0\quad y(0)=1; y'(0)=1.

1. What is the meaning of the notation y''+y'-2y=0?

The notation represents a second-order linear differential equation, where y'' represents the second derivative of y with respect to some independent variable, y' represents the first derivative of y, and y represents the dependent variable.

2. How do you find the solution of a second-order differential equation?

To find the solution of a second-order differential equation, you need to use a method called "undetermined coefficients." This involves assuming a solution of the form y = e^rx, where r is a constant, and solving for the value of r that satisfies the equation.

3. What does y(0)=1 and y'(0)=1 mean in this context?

These initial conditions represent the values of y and its derivative at the starting point, which in this case is when the independent variable is equal to 0. This information is necessary to uniquely determine the solution to the differential equation.

4. How do you solve a differential equation with initial conditions?

First, you need to find the general solution to the differential equation. Then, you can use the initial conditions to determine the values of any arbitrary constants in the general solution. This will give you the specific solution that satisfies the given initial conditions.

5. What is the significance of finding the solution to a differential equation?

Finding the solution to a differential equation allows us to model and understand various phenomena in the natural world, such as population growth, motion, and electrical circuits. It also has applications in many fields, including physics, engineering, and economics.

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