# Find the function y_1(t) which is the solution of 9y'' - 24y' + 7y = 0 with initial...

## Question:

Find the function {eq}y_1(t) {/eq} which is the solution of {eq}9y'' - 24y' + 7y = 0 {/eq} with initial conditions {eq}y_1 (0) = 1, y'_1(0) = 0 {/eq}.

## Initial value problem:

Linear differential equation with constant coefficients:

The general form of this equation is

{eq}\displaystyle \frac{d^ny}{dt^n}+a_1\frac{d^{n-1}y}{dt^{n-1}}+a_2\frac{d^{n-2}y}{dt^{n-2}}+...+a_ny=0 {/eq}, where {eq}\displaystyle a_1, a_2,\cdots, a_n {/eq} are constants.

To find the general solution of {eq}\displaystyle (D^n+a_1D^{n-1}+a_2D^{n-2}+\cdots+a_n)y=0..............(1) {/eq}, (1) is equivalent to {eq}\displaystyle (D-m_1)(D-m_2)\cdots(D-m_n)y=0..................................(2) {/eq}.

Then solutions of any one of the equations {eq}\displaystyle (D-m_1)y=0,~~~ (D-m_2)y=0,~~~ ...,~~~(D-m_n)y=0 {/eq} is also a solution of (2).

Taking {eq}\displaystyle y=e^{mt} {/eq} we can find the auxiliary equation (A.E) of (2). So (1) becomes {eq}\displaystyle (m^n+a_1m^{n-1}+a_2m^{n-2}+...+a_n)e^{mt}=0\\ \Rightarrow (m^n+a_1m^{n-1}+a_2m^{n-2}+...+a_n)=0.............(3). {/eq} When the A.E (3) is of degree 2 (n=2) with two real and different roots {eq}m_1{/eq} and {eq}m_2 {/eq} then the general solution is {eq}\displaystyle y(t)=c_1 e^{m_1 t}+c_2 e^{m_2 t} {/eq}.

## Answer and Explanation:

The given equation is {eq}\displaystyle 9y'' - 24y' + 7y = 0.......(1) {/eq}.

The A.E of (1) is: {eq}9m^2-24m+7=0 \\ \Rightarrow 9m^2-21m-3m+7=0 \Rightarrow 3m(3m-7)-(3m-7)=0 \Rightarrow (3m-7)(3m-1)=0 \\ \Rightarrow m=\frac{7}{3}, ~or~ m=\frac{1}{3}.........(2) {/eq}

Then from (2), we get {eq}m_1=\frac{7}{3}, m_2=\frac{1}{3} {/eq}.

Hence, {eq}\displaystyle y_1 (t)=c_1 e^{\frac{7t}{3}}+c_2 e^{\frac{t}{3}}......(3) {/eq}.

Now, {eq}\displaystyle y_1'(t)=c_1 \frac{7}{3}e^{\frac{7t}{3}}+c_2 \frac{1}{3}e^{\frac{t}{3}}..........(4) {/eq}.

Given {eq}\displaystyle y_1(0) = 1, y_1'(0) = 0.......(5) {/eq}.

Therefore, {eq}\displaystyle y_1(0) = 1 \Rightarrow c_1+c_2=1..............(6) {/eq},

and {eq}\displaystyle y'(0) = 0 \Rightarrow \frac{7 c_1}{3}+\frac{c_2}{3}=0 {/eq} {eq}\displaystyle \Rightarrow 7c_1+c_2=0.............(7) {/eq}.

Solving equation (6) and (7), we get {eq}\displaystyle c_1=-\frac{1}{6} ~ and~ c_2=\frac{7}{6} {/eq}.

Hence, from (3), we get the function is {eq}\displaystyle y_1(t)=-\frac{1}{6}e^{\frac{7t}{3}}+\frac{7}{6}e^{\frac{t}{3}} {/eq}.

#### Ask a question

Our experts can answer your tough homework and study questions.

Ask a question Ask a question#### Search Answers

#### Learn more about this topic:

from Calculus: Help and Review

Chapter 11 / Lesson 13