# Thread: 2nd ODE -- Finding Particular solution with A constant term !!

1. ## 2nd ODE -- Finding Particular solution with A constant term !!

Hello all,

I am struggling to solve this question ..
I know how to solve non-homogenous 2nd ODE if we have cos, sin and exponents .. but for in this question, we have a constant term.
I replaced the constant term by k for the particular solution .. is it wrong ?

A second order differential equation is given by 3.7(d2x/dt2t +3dx/dt+2x)=0.8, with initial condition x(0)=-1 and dx/dt(0)=-2. Solve the ODE and hence compute x(2.9). Give the answer to 3 decimal places.

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How do I get that answer ? what am I doing wrong ?

2. For your particular solution try a constant: For instance if your equation is $x''+ax'+bx=c$ with $a,b,c$ constant then if $c\neq 0$ we have that $p=\frac{b}{c}$ is a solution. If you want to you can think of this as a trivial application of the indeterminate coefficients method.

Yes thats the answer for the particular solution, but what about the rest ?

Is the general solution = y=C1e^-2t+C2e^-t +37/4 ?

Kind Regards,
Z

4. Originally Posted by ZaZu
Is the general solution = y=C1e^-2t+C2e^-t +37/4 ?

Indeed. Notice that your eqation is equivalent to $x''+3x'+2x= \frac{.8}{3.7}$ And with the argument I gave you in the other post find the particular solution.

For the rest you have $x(t)=c_1e^{-2t} + c_2e^{-t} + x_p$ where $x_p$ is your part. sol. Now given the initial conditions you have to determine $c_1$ and $c_2$ (It's easy although a bit tedious, you'll end with a linear system of two equations with two variables $c_1$ and $c_2$) and once you have them just evaluate.

5. im sorry but im still not getting the right answer ..

This is the first time im doing 1st and 2nd order differentials so its kind of new to me.

I will go watch some educational youtube videos about the subject and hopefully ill understand ..

If you can please explain this more thoroughly, then please do, Im not forcing you or anything, you decide to if you want to or not

Thanks,
Z