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How do I solve $y''+y'+7y=t$ where $y(0)=0$ and $y'(0)=0$ $(t\geq 0)$?

I tried to solve this by Laplace transformation, but I couldn't find the inverse of $1/(s^2(s^2+s+7))$.

How would I solve this?

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    $\begingroup$ Is $t$ a constant? $\endgroup$
    – wythagoras
    Jun 12 '15 at 7:43
  • $\begingroup$ @wythagoras No, it's a variable and $y=y(t)$ $\endgroup$
    – Rubertos
    Jun 12 '15 at 7:45
  • $\begingroup$ Try doing a partial fraction decomposition on the Laplace transform $\endgroup$
    – MrSlunk
    Jun 12 '15 at 8:15
  • $\begingroup$ @MrSlunk Would you please help me in details $\endgroup$
    – Rubertos
    Jun 12 '15 at 8:38
  • $\begingroup$ Partial fraction decomposition of $1/s^2(s^2+s+7))$ is really really dirty including complex numbers $\endgroup$
    – Rubertos
    Jun 12 '15 at 8:39
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You have the Laplace transform, so lets do a partial fraction decomposition. We want to find $a,b,c,d$ such that $$\frac{1}{s^2(s^2 + s +7)} = \frac{a}{s} + \frac{b}{s^2} + \frac{cs + d}{s^2 + s + 7}.$$ So we have $$as(s^2 + s +7) + b(s^2 + s +7) + cs^3 + ds^2 = 1.$$ Collecting coefficients we have $$ (a+c)s^3 + (a+b+d)s^2 + (7a+b)s + 7b =1$$ So after working your way through this, you should find, $$a = -\frac{1}{49},b = \frac{1}{7}, d= -\frac{6}{49}, c = \frac{1}{49}.$$ The final trick is to complete the square so that you have $$ s^2+s +7 = \left(s+\frac{1}{2}\right)^2 +\left(\frac{3\sqrt{3}}{2}\right)^2.$$ So $$\frac{1}{s^2(s^2 + s +7)} = \frac{1}{49}\left(\frac{-1}{s} + \frac{7}{s^2} + \frac{s+\frac{1}{2} }{\left(s+\frac{1}{2}\right)^2 +\frac{27}{4}}+\frac{13}{3\sqrt{3}}\frac{\frac{3\sqrt{3}}{2}}{\left(s+\frac{1}{2}\right)^2 +\frac{27}{4}}\right).$$ From this point the solution can be read off tables.

Edit:

I'd like to add some insight into wythagoras' answer.

Let's suppose you have some function $y$. Suppose then you differentiate it a couple of times and you end up with a polynomial. Then it follows that $y$ had to be a polynomial to start with. Remember that differentiating polynomials gives you another polynomial of one less order. Hence, if i add a polynomial to it's derivatives, then what comes out the other end must be the same order as what i started with. So when i have $y'' + y'+7y =t$ then i know that i must have a polynomial, and that the highest order of that polynomial must be of order $t$. So we are well justified assuming that the particular solution has the form $y_p= bt +a$.

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Hint: The charactaristic equation is $$1+x+7x^2=0$$

Solve it. You will find two complex roots.


Then find a particular solution.

A hint for finding it:

  • What happens when you put $y=\frac{1}{7}t$?
  • How can you prevent that?

Method 1. Substiute $at+b$ to get $7at+7b+a=t$. Thus $7a=1$, thus $a=\frac{1}{7}$. Then $7b+a=0$, thus $b=-\frac{1}{49}$.

Therefore the particular solution is $\frac{1}{7}t-\frac{1}{49}$

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  • $\begingroup$ This is what I exactly tried but I am still stuck. I solved this by using wronskian , but I think there is another way to do this $\endgroup$
    – Rubertos
    Jun 12 '15 at 7:53
  • $\begingroup$ Another hint: When you differentiate a constant, it will become zero. So if you add a constant to $y$, the derivatives don't change. $\endgroup$
    – wythagoras
    Jun 12 '15 at 7:54
  • $\begingroup$ So is this equation generally solved by judicious guessing? $\endgroup$
    – Rubertos
    Jun 12 '15 at 8:01
  • $\begingroup$ Uh... I wouldn't call it guessing. Usually when you want to find a particular solution you substitute in $y=f(t)$, with $f$ the same form as the RHS. So in this case, you could subsitute in $at+b$, and solve for $a,b$. I will work it out in my answer. $\endgroup$
    – wythagoras
    Jun 12 '15 at 8:04
  • $\begingroup$ I added it to my answer. Now you can have the general solution in the form $Ae^{a_1t}+Be^{a_2t}+\frac{1}{7}t-\frac{1}{7}$, then solve for $A,B$ using your initial conditions. $\endgroup$
    – wythagoras
    Jun 12 '15 at 8:10
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The homogeneous equation shouldn't be a problem to you.

For a particular solution, notice that the derivative of the RHS is a constant, and the second derivative vanishes. So plugging $y=t$ you get

$$0+1+7t=t$$ not so far from the solution. You fix by adjusting the coefficient of $t$ to $t/7$ and adding a constant to compensate the $1$:

$$0+\frac17+7(\frac t7+c)=t,$$then $$\frac t7-\frac1{49}.$$

This trial-and-error process (an informal application of the indeterminate coefficients method) will work with polynomials and exponentials. First try $y=RHS(t)$, observe the patterns and fix the discrepancies.

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