Answer to Question #190654 in Differential Equations for Didues

Question

y"+3y=0 , y(0)=0 and y'(3)=0

Answer 1

Given, y''+3y=0,y(0)=0 and y'(3)=0

$\Rightarrow y''+3y=0$

$(D^2+3)y=0$

So Auxiliary equation-

$m^2+3=0\Rightarrow m=\pm \sqrt{3}i$

Solution of the equation is-

$y=c_1cos\sqrt{3}x+c_2sin\sqrt{3}x~~~~~-(1)$

At x=0,y=0

$0=c_1cos0+c_2sin 0\Rightarrow c_1=0$

Differentiating equation (1) w.r.t. x-

$y'=-\sqrt{3}c_1sin\sqrt{3}x+\sqrt{3}c_2cos\sqrt{3}x$

Also, $y'(3)=0$

$\Rightarrow 0=-\sqrt{3}c_1sin3\sqrt{3}+\sqrt{3}c_2 cos3\sqrt{3}$

As $c_1=0\Rightarrow c_2=0$

Solution of equation is

y=0