How to solve this second order diff eq, need help!?
2500y''+9y=0;
y(0) = 1
y'(0) = 1
I think you have to substitute with z = dy/dt and reduce it to a separable equation..but somehow the math is still not working out for me.
Hello, cheeee!
This is not a "separable" problem . . .
The characteristic equation is: .$\displaystyle 2500m^2 + 9 \:=\:0$Solve: .$\displaystyle 2500y''+9y\:=\:0,\quad
y(0) \,= \,1,\;\;y'(0) \,=\, 1$
. . which has roots: .$\displaystyle m \:=\:\pm\frac{3}{50}i $
Hence, the general solution is: .$\displaystyle y \;=\;C_1\cos\left(\frac{3}{50}x\right) + C_2\sin\left(\frac{3}{50}x\right) $
Since $\displaystyle y(0) = 1\!:\;\;C_1\cos(0) + C_2\sin(0) \:=\:1\quad\Rightarrow\quad\boxed{ C_1 \:=\:1}$
Differentiate: .$\displaystyle y' \;=\;-\frac{3}{50}C_1\sin\left(\frac{3}{50}x\right) + \frac{3}{50}C_2\cos\left(\frac{3}{50}x\right) $
Since $\displaystyle y'(0)=1\!:\;\;-\frac{3}{50}C_1\sin(0) + \frac{3}{50}C_2\cos(0) \:=\:1\quad\Rightarrow\quad\boxed{ C_2 = \frac{50}{3}}$
Therefore: . $\displaystyle \boxed{\;y \;=\;\cos\left(\frac{3}{50}x\right) + \frac{50}{3}\sin\left(\frac{3}{50}x\right)\;} $
$\displaystyle 50^2y''+3^2y=0$
solve the equation
$\displaystyle 50^2m^2+3^2=0$
rewriting we get
$\displaystyle (50m)^2+3^2=(50m)^2-(3i)^2=(50m-3i)(50m+3i)=0$
so the two solutions for m are
$\displaystyle m=\frac{3i}{50}$ or $\displaystyle =\frac{-3i}{50}$
so we end up with the complimentry solution.
$\displaystyle y_c=c_1\cos(\frac{3}{50}t)+c_2\sin{\frac{3}{50}t}$