MAT244--2019F > Quiz-3

TUT0602 QUIZ3

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**Kun Zheng**:

Hi everyone, my question is to get the solution of y"-4y'=0, use the initial points of y'(-2)=1 and y(-2)=-1

First of all, we assume that y=e^(rt), and r must be a root of the characteristic equation.

Hence, we rewrite it as:

$r^2 -4r=0$

$r(r-4)=0$

$r_1=0, r_2=4$

Then we have the general structure as:

$y=C_1e^{r_1t}+C_2e^{r_1t}$

$y=C_1+C_2e^{4t}$

Derivative $y=C_1+C_2e^{4t}$

$y'=4C_2e^{4t}$

Use the initial values to plug in y'(-2)=1, y(-2)=-1

Got $C_2=e^8/4$

Then $y=C_1+e^{4t+8}/4$

Got $C_1=3/4$

Therefore, the initial equation is $y=3/4+e^{4t+8}/4$

Note: $y\rightarrow \infty, t\rightarrow \infty$

Correct me if I made a wrong solution or wrong question!

Have a good weekend!

**Vickyyy**:

hi, is the question y''-4y'=0 instead of y''-4y''=0?

**Kun Zheng**:

--- Quote from: Vickyyy on October 13, 2019, 12:57:42 AM ---hi, is the question y''-4y'=0 instead of y''-4y''=0?

--- End quote ---

Thank you! My mistake.

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