$$4y''-4y'+y=0$$
y(0)=1 y'(0)=-1.5
auxiliary equation
$$4r^2-4r+=0$$
$$(2r-1)(2r-1)=0$$
$$r_1=.5$$
$$r_2=.5$$
$$y(x)=c_1e^{.5x}+c_2e^{.5x}$$
$$y(0)=c_1e^{.5x}+c_2e^{.5x}=0$$
$$y'(0)=c_10.5e^{0.5x}+c_20.5e^{0.5x}=-1.5$$

Question

$$4y''-4y'+y=0$$
y(0)=1       y'(0)=-1.5
auxiliary equation
$$4r^2-4r+=0$$
$$(2r-1)(2r-1)=0$$
$$r_1=.5$$
$$r_2=.5$$
$$y(x)=c_1e^{.5x}+c_2e^{.5x}$$
$$y(0)=c_1e^{.5x}+c_2e^{.5x}=0$$
$$y'(0)=c_10.5e^{0.5x}+c_20.5e^{0.5x}=-1.5$$

turingtest · Answer

so you are having trouble finding c1 and c2?

lgbasallote · Answer

wont it be better to use 1/2 instead of 0.5?

anonymous · Answer

yeah :S

lgbasallote · Answer

..just saying

turingtest · Answer

you did not plug in x=0 in you particulars

anonymous · Answer

forgot to do that

lgbasallote · Answer

you also equated y(0) to 0

lgbasallote · Answer

should be y(0) = 1

anonymous · Answer

$$y(0)=c_1e^{.5(0)}+c_2e^{.5(0)}=1$$
$$y(0)=c_1+c_2=1$$

turingtest · Answer

$$y(0)=c_1e^{0}+c_2e^{0}=c_1+c_2=1$$$$y'(0)=0.5c_10+0.5c_2=-1.5$$

anonymous · Answer

repeated roots mean the solution looks like  C1*e^lambda*x + C2*x*e^lambda*x

turingtest · Answer

good call, I missed that

turingtest · Answer

so$$y(0)=c_1e^{0}+c_2(0)e^{0}=c_1=1$$$$y'(0)=0.5(1)+0.5c_2=-1.5$$now it's easy

anonymous · Answer

why is $$c_2(0)e^0$$ why do you have it multiplied by zero

turingtest · Answer

because of the repeated root you should have had$$y(x)=c_1e^{.5x}+c_2xe^{.5x}$$

turingtest · Answer

from which the above follows

anonymous · Answer

oh I see.

anonymous · Answer

and now it's just a matter of substitution
Thanks!!!!

turingtest · Answer

welcome!

anonymous · Answer

No problem.