Question

If $y\text{'}\text{'}-3y\text{'}+2y=0$ where $y\left(0\right)=1,y\text{'}\left(0\right)=0$, then the value of $y$ at $x=\log 2$ is

• A. $1$
• B. $-1$
• C. $2$
• D. $0$

Answer 1

The correct option is D

$0$

Explanation for the correct option.

Step 1. Find the general equation of $y$.

It is given that $y\text{'}\text{'}-3y\text{'}+2y=0$, so $\frac{d^{2}y}{dx}-3\frac{dy}{dx}+2y=0$.

The corresponding equation is: $m^2-3m+2=0$.

Let the general solution be given by the equation $y=A{e}^x+B{e}^{2x}...\left(1\right)$.

Now find the value of $A$ and $B$.

It is given that at $x=0$, the value of $y$ is $1$. So substitute $0$ for $x$ and $1$ for $y$ in the equation $1$.

$$\begin{gathered} \begin{array}{rcl}1& =& A{e}^0+B{e}^{2\times 0}\\ & \Rightarrow & 1=A\times 1+B\times 1\end{array} \\ \begin{array}{rcl}& \Rightarrow & 1=A+B...\left(2\right)\end{array} \end{gathered}$$

Now differentiate equation $1$, $y=A{e}^x+B{e}^{2x}$ with respect to $x$.

$$\begin{gathered} y\text{'}=A{e}^x+B{e}^{2x}\times 2 \\ \Rightarrow y\text{'}=A{e}^x+2B{e}^{2x}...\left(3\right) \end{gathered}$$

It is given that at $x=0$, the value of $y\text{'}$ is $0$. So substitute $0$ for $x$ and $0$ for $y\text{'}$ in the equation $3$.

$$\begin{gathered} 0=A{e}^0+2B{e}^{2\times 0} \\ \Rightarrow 0=A\times 1+2B\times 1 \\ \Rightarrow 0=A+2B...\left(4\right) \end{gathered}$$

Now subtract equation $2$ from equation $4$.

$$\begin{gathered} 0-1=(A+2B)-(A+B) \\ \Rightarrow -1=A+2B-A-B \\ \Rightarrow -1=B \end{gathered}$$

Now substitute $-1$ for $B$ in equation $2$.

$$\begin{gathered} 1=A+(-1) \\ \Rightarrow 1=A-1 \\ \Rightarrow 1+1=A \\ \Rightarrow 2=A \end{gathered}$$

So in the general equation $y=A{e}^x+B{e}^{2x}$ substitute $2$ for $A$ and $-1$ for $B$ to get the solution of the differential equation.

$$\begin{gathered} y=2e^x+(-1)e^{2x} \\ \Rightarrow y=2e^x-e^{2x} \end{gathered}$$

Step 2. Find the value of $y$ at $x=\log 2$.

In the equation $y=2e^x-e^{2x}$ substitute $\log 2$ for $x$ and find the value of $y$ using properties of logarithm.

$\begin{array}{rcl}y& =& 2e^{\log 2}-e^{2\log 2}\\ & =& 2e^{\log 2}-e^{\log \left(2^2\right)}\left[m\log a=\log \left(a^m\right)\right]\\ & =& 2\times 2-2^2\left[b^{{\log }_b\left(a\right)}=a\right]\\ & =& 4-4\\ & =& 0\end{array}$

So, the value of $y$ at $x=\log 2$ is $0$.

Hence, the correct option is D.