Solving Differential Equations

[CatchThis2]

Solve the differential equation with given initial condition: dy/dt= 0.5y; y(2)=100

So far I took the derivative of .5y and got (1/4)y^2 and added +c=100

Now I have (1/4)(2)^2+c=100

Y=1+c=100
Y=99

I get the equation Y=(1/4)x^2+99 but for some reason that answer is wrong.

Does anyone see where I am going wrong?

 

[swaroop]

dy/dt=0.5*y
=> dy/y=0.5*dt
Integ both sides
ln(y)=0.5t+constant
Since y=100 at t=2,
ln(100)=1+constant
=>constant=3.605
Hence the equation is
y=exp(0.5t+3.605)

 

[CatchThis2]

Thanks for the help, I appreciate it!

 

[BigGlenntheHeavy]

\(\displaystyle Given: \ \frac{dy}{dt} \ = \ .5y \ and \ y(2) \ = \ 100. \ First \ note \ y \ = \ 0 \ is \ a \ solution.\)

\(\displaystyle To \ find \ other \ solutions, \ we \ assume \ y \ \ne \ 0.\)

\(\displaystyle Hence, \ \int\frac{dy}{y} \ = \ \int.5dt \ \implies \ ln|y| \ = \ .5t+C\)

\(\displaystyle Ergo, \ |y| \ = \ (e^{C})(e^{.5t}), \ y \ = \ \pm(e^{C})(e^{.5t})\)

\(\displaystyle Because \ y \ = \ 0 \ is \ also \ a \ solution, \ we \ can \ write \ the \ general \ solution \ as \ y(t) \ = \ Ae^{.5t}, \ A \ = \ \pm e^{C}\)

\(\displaystyle \ or \ A \ = \ 0.\)

\(\displaystyle Then, \ y(2) \ = \ 100 \ = \ Ae, \ A \ = \ \frac{100}{e}\)

\(\displaystyle Therefore, \ y(t) \ = \ \bigg(\frac{100}{e}\bigg)e^{.5t} \ = \ 100e^{.5t-1}, \ y \ \ne \ 0.\)