Particular Solution to Differential Equation

[The_Flying_Skeptic]

5dy/dt + 10y(t) = 2x(t)

x(t) = 2

so would a particular solution be y(t)=0.4 ?

[Angrboda]

[TheRealJoeFish]

5dy/dt + 10y(t) = 2x(t)

x(t) = 2

so would a particular solution be y(t)=0.4 ?

Oooooh boy it's been a while since my DiffEq days.  I think it would be because, if y(t) is the constant .4, its derivative is 0, so the first term would be 5*0, the second term would be 10*.4, and the sum would be 2*2, for 0+4=4.

I think that's how maths work, but someone feel free to correct me otherwise. I'm not sure if you can just do it like that.

[houseofcantor]

Wolfram Alpha says 2.
I think alpha is saying you done fucked up.

(Yeah, it's only been 15 years.) 

[robvalue]

Been a while here too, haha!

If x(t) = 2 so it's just a constant, that gives

5y' + 10y = 4

So the particular solution would be found by y=c as higher powers of t would end up with zero coefficient, giving

5*0 + 10*c = 4

Then c = 0.4

Solving the homogenous equation

5y' + 10y = 0

y' + 2y = 0

Needs multiplying by the integrating factor of e^(integral of coefficient of y) = e^(2t)

y'e^(2t) + 2ye^(2t) = 0

(ye^(2t))' = 0

ye^(2t) = A

y = Ae^(-2t)

So complete solution is y = Ae^(-2t) + 0.4

If my memory serves me correctly!

[Alex K]

Looks good, rob

[robvalue]

Thanks! A bit of rust on the old cogs, took some dislodging

[Angrboda]

I'm still confoozled. I never made it through DiffEq.

[robvalue]

I can explain it in more detail, if you like. How much do you know about calculus?

[GrandizerII]

I'm still confoozled.  I never made it through DiffEq.

Probably due to lack of interest rather than anything else. You're obviously very intelligent as shown by your posts, so you can learn it well enough with practice and dedication. Sequential learning and all that.

I used to think I could never be good at such "advanced" maths back in high school, but thanks to the Internet, it's become as easy as driving for me.  I had terrible high school teachers, no doubt.