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$$f(t) - f''(t) = \frac{\pi}{2e^t}$$

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I would have been able to solve it if the right side was a constant.

Can someone please fully solve it. I feel really dumb right now..I've been stuck on it for a very long time

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  • $\begingroup$ I don't get the logic of why we're substituting it as $e^{\lambda t}$ $\endgroup$
    – Rick
    Dec 30, 2017 at 13:11

7 Answers 7

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Take $g(t)=f(t)-f'(t)$ and get $g(t)+g'(t)=\frac{\pi}{2e^t}$ or $$\frac d{dt}(e^tg(t))=\frac{\pi}2$$ That is $f(t)-f'(t)=\frac{\pi t}{2e^t}$ which implies $$\frac d{dt}(-e^{-t}f(t))=\frac {\pi t}{2e^{2t}}$$ or $$f(t)=-\frac{\pi}2e^t\int te^{-2t}dt+c$$

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  • $\begingroup$ I'll get $f(t) - f'(t) = \frac{\pi t}{2e^t}$ $\endgroup$
    – Rick
    Dec 30, 2017 at 13:46
  • $\begingroup$ $\frac{d}{dt}(-e^{-t}f(t)) =e^{-t}(f(t)-f'(t))$ $\endgroup$
    – QED
    Dec 30, 2017 at 13:49
  • $\begingroup$ Can you please fully solve it :) I've been stuck for a really long time $\endgroup$
    – Rick
    Dec 30, 2017 at 14:15
  • $\begingroup$ Thank you, can you tell me why we ignored the integration constant in the second step? it gives a diff result if we consider it. $\endgroup$
    – Rick
    Dec 31, 2017 at 11:26
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    $\begingroup$ Happy new year to you too! Cheers! $\endgroup$
    – QED
    Dec 31, 2017 at 17:22
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Hint: let $y=ze^{-t}$. Which differential equation does $z$ satisfy?

Why this choice:

We see that the term on the right hand side is of the form $g(t)e^{-t}$, where $g(t)$ is a constant in this case. What I want to do is get rid of this $e^{-t}$ term so that I will get an easy differential equation. But I know that $(e^{-t})'=-e^{-t}$, so I know that no matter how many times I differentiate any function of the form $ze^{-t}$, I will still get a factor $e^{-t}$, so that I will be able to divide both the right hand side and the left hand side by this term, which will lead to an easy to solve differential equation.

Note: it is only a coincidence in this case that the right hand side is of the form $e^{\lambda t}$. In general, with the method I gave, you can solve the equation for any power of $e$. So it would work for $e^{3t}$ or $e^{-5t}$. There are shortcuts for the case $e^{\lambda t}$, though.

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Hint:

The r.h.s. has the form $C\mathrm e^{-t}$, and $-1$ is a root of the characteristic equation. Try a particular solution of the form: $$y_0(t)=Ct\,\mathrm e^{-t}, \quad C\in\mathbf R. $$

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Hint:

Substitute $g(t)=f(t)+f'(t)$

then your equation become a first order equation:

$$f(t) +f'(t)-f'(t)-f''(t)=h(t)$$ $$g(t)-g'(t)=h(t)$$

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It's good that you don't get the logic of substituting it as $e^{\lambda t} $. The right approach is not learning such cheap tricks but to solve the equation step by step.

If we use the symbol $D$ to denote differentiation with respect to $t$ and powers of $D$ to denote repeated differentiation so that $$Df(t) =\frac{d} {dt} f(t), D^{n} f(t) =\frac{d^{n} } {dt^{n}} f(t) $$ then we can think of polynomials in $D$ with constant coefficients so that $$(a_{0}D^{n}+a_{1}D^{n-1}+\dots+a_{n-1}D+a_n)f(t)=\\ a_{0}f^{(n)}(t)+a_{1}f^{(n-1)}(t)+\dots+a_{n-1}f'(t) +a_{n} f(t) $$ This way of thinking has certain advantages in the sense that we can then add / multiply such polynomials according to the usual rules of operations with polynomials (this needs to be proved by the way and its not difficult to do so).


The equation in your question can then be written as $$(1-D^2)f(t)=(\pi/2)e^{-t}\tag{1}$$ The right way to solve such equations is to understand that a linear equation of type $$(D+a)p(t)=q(t)\tag{2}$$ can always be solved by multiplying the above equation with $e^{at} $. Note that $$D(e^{at} p(t)) = e^{at} Dp(t) +a e^{at} p(t) =e^{at} (D+a)p(t) $$ and hence multiplying equation $(2)$ by $e^{at} $ gives us $$D(e^{at}p(t))= e^{at} q(t) $$ Integrating this equation with respect to $t$ gives us $e^{at} p(t) $ and hence $p(t) $. Thus we need to solve $(1)$ by reducing it to a series of equations of type $(2)$. How? Just by factorizing the polynomial occurring in the equation $(1)$.


Thus we rewrite $(1)$ as $$(1-D)(1+D)f(t)=(\pi/2)e^{-t}$$ and set $$(1+D)f(t)=g(t)\tag{3}$$ so that $$(1-D)g(t)=(\pi/2)e^{-t} \tag{4}$$ The given equation $(1)$ is thus reduced to a set of two linear equations $(3),(4)$ which are easily solved in succession as described earlier. Thus multiplying $(4)$ by $e^{-t}$ we get $$-D(e^{-t} g(t)) =(\pi/2)e^{-2t}$$ On integrating we get $$e^{-t} g(t) =\frac{\pi} {4}e^{-2t}+c$$ or $$g(t) =ce^{t} +\frac{\pi} {4}e^{-t}$$ Equation $(3)$ can now be solved in similar manner by multiplying it with $e^{t} $ and we then have $$D(e^{t} f(t)) = ce^{2t}+\frac{\pi}{4}$$ On integrating we get $$e^{t} f(t) =\frac{ce^{2t}}{2}+\frac{\pi t} {4}+d$$ or $$f(t) =Ae^{t} +Be^{-t}+\frac{\pi} {4}te^{-t}$$ where $A=c/2, B=d$ are arbitrary constants.


It should be noted that the above procedure involves writing the equation using differential operator $D$ and factoring the polynomial in $D$ (also known as characteristic polynomial of the equation) down to the linear factors and handling each factor one by one.

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Hint: Use characteristic equation $1-\lambda^2=0$ and find the general solution, then with $y=Ate^{-t}$ find particular solution.

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  • $\begingroup$ Happy new year :) $\endgroup$
    – Nosrati
    Dec 30, 2017 at 15:00
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I use for simplicity $K=\frac {\pi} 2$

$$f(t) +f'(t)-f'(t)-f''(t)=Ke^ {-t}$$ Multiply by $e^t$ $$f(t)e^t +f'(t)e^t-(f'(t)e^t+f''(t)e^t)=K$$

$$(f(t)e^t)'-(f'(t)e^t)'=K$$

integrate both side $$(f(t)e^t)-(f'(t)e^t)=Kt+C$$ Divide by $e^{2t}$ $$\frac {f(t)e^t-f'(t)e^t}{e^{2t}}=e^{-2t}(Kt+C)$$

Then,

$$-(\frac {f(t)}{e^{t}})'=e^{-2t}(Kt+C)$$ Integrate again, $$f(t)=-e^{t}\int e^{-2t}(Kt+C)dt$$

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