Solving y' - (1/2)y = 2cos(t) for y

[Pantha]

Hi! I just fished a math project and got the grade on it but the one question I got wrong is bothering me I must know the answer can anyone help me? here s that question.

Solve the following for y:

y' - (1/2)y = 2cos(t)

Also, please show me how you got the answer.

 

[Denis]

Is that 1/ (2y) OR (1/2)y ?

Can you solve this for y: x - (3/4)y = 5z ?

 

[Pantha]

Its (1/2)y sec and I'll have a answer for you problem I'm a dyslexic female. This makes it hard for me to do math but I get it done. XD

 

[stapel]

Its (1/2)y sec...

So the equation is actually as follows?

. . . . .y' - (1/2)y sec(t) = 2cos(t)

You've posted this to "Geometry and Trig", but the y-prime makes this look like some sort of differential equation. How did this equation arise? What relationship do you have between y and t?

Please be specific. Thank you.

Eliz.

 

[Pantha]

That’s how it was given to me. -_- I don't get it ether. I was hoping one of could help.

 

[Pantha]

A friend just told me that y is the prime

 

[galactus]

This is a first order linear equation.

\(\displaystyle \L\\\frac{dy}{dt}-\frac{y}{2}=2cos(t)\)

Integrating factor would be \(\displaystyle \L\\e^{\frac{-t}{2}}\)


\(\displaystyle \L\\\frac{d}{dt}[ye^{\frac{-t}{2}}]=2e^{\frac{-t}{2}}cos(t)\)

Integrate both sides(the right side can be done using 'parts'. I'll leave that for you to struggle with):

\(\displaystyle \L\\ye^{\frac{-t}{2}}=\frac{-4(cos(t)-2sin(t))e^{\frac{-t}{2}}}{5}+C\)

\(\displaystyle \L\\y=\frac{-4}{5}cos(t)+\frac{8}{5}sin(t)+Ce^{\frac{t}{2}}\)

 

[stapel]

That’s how it was given to me. -_- I don't get it ether. I was hoping one of could help.

Have you not taken a "differential equations" course yet...?

Thank you.

Eliz.