Question

Solve the initial-value problem. y2y3y=0, y(0)=2, y(0)=2y^{\prime \prime}-2 y^{\prime}-3 y=0, \ y(0)=2, \ y^{\prime}(0)=2

Solution

Verified
Step 1
1 of 3

It is given that:

y2y3y=0y(0)=2y(0)=2\begin{align*} y^{\prime\prime}-2y^\prime-3y&=0\\ y(0)&=2\\ y^\prime(0)&=2 \end{align*}

In order to solve the initial-value problem we will first solve the given differential equation.,The auxiliary equation is:

r22r3=0r^2-2r-3=0

Now,let us solve this quadratic equation:

r1,2=2±(2)241(3)2=2±4+122=2±162=2±42r1=3,r2=1\begin{align*} r_{1,2}&=\dfrac{2 \pm \sqrt{(-2)^2-4\cdot 1 \cdot (-3)}}{2}\\ &=\dfrac{2\pm \sqrt{4+12}}{2}\\ &=\dfrac{2\pm \sqrt{16}}{2}\\ &=\dfrac{2\pm 4}{2}\Rightarrow \pmb{r_1=3\,\, , \,\, r_2=-1} \end{align*}

Notice that this is the case when b24ac>0b^2-4ac>0 so the general solution has the form

y(x)=c1er1x+c2er2xy(x)=c_1 e^{r_1x}+c_2 e^{r_2x}

Substituting r1r_1 and r2r_2 we conclude that the general solution of the given differential equation is

y(x)=c1e3x+c2ex\boxed{y(x)=c_1 e^{3x}+c_2 e^{-x}}

Create an account to view solutions

Create an account to view solutions

Recommended textbook solutions

Calculus: Early Transcendentals 7th Edition by James Stewart

Calculus: Early Transcendentals

7th EditionISBN: 9780538497909James Stewart
10,082 solutions
Calculus: Early Transcendentals 8th Edition by James Stewart

Calculus: Early Transcendentals

8th EditionISBN: 9781285741550 (6 more)James Stewart
11,083 solutions
Calculus: Early Transcendentals 9th Edition by Daniel K. Clegg, James Stewart, Saleem Watson

Calculus: Early Transcendentals

9th EditionISBN: 9781337613927 (4 more)Daniel K. Clegg, James Stewart, Saleem Watson
11,050 solutions
The Practice of Statistics for the AP Exam 5th Edition by Daniel S. Yates, Daren S. Starnes, David Moore, Josh Tabor

The Practice of Statistics for the AP Exam

5th EditionISBN: 9781464108730 (2 more)Daniel S. Yates, Daren S. Starnes, David Moore, Josh Tabor
2,433 solutions

More related questions

1/4

1/7