I can solve a second order ODE, but I am stuck at this. Is this reducible to something solvable : $({dy\over dx}-1)^2({d^2y \over dx^2} +1)^2y = sin^2{x \over 2}+e^x+x$ ?
Solve $({dy\over dx}-1)^2({d^2y \over dx^2} +1)^2y = sin^2{x \over 2}+e^x+x$
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$\begingroup$
calculus
ordinary-differential-equations
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0Should the left-hand side of this be $(D-1)^2(D^2+1)y$ with $D$ representing differentiation by $x$? That would make more sense in terms of your apparent existing knowledge. In any case note that $\cos x=1-2\sin^2\frac x2$. – 2017-02-28
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0Yes @mark, I can use the differential operator D to write the equation. LHS of equation is $(Dy-1)^2(D^2y+1)^2y$. Then, should the equation be treated as a polynomial equation in D ? Actually I use D very sparingly coz I am a little uncomfortable with that. Can you provide the next step as per your understanding. – 2017-02-28
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0@kloikfriend: both your expressions seem unlikely (with $d/dx$ and with $D$). Double check the problem statement, I don't think you reproduced it faithfully. – 2017-02-28
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0@YvesDaoust my D notation may be wrong (I dont know). But the statement in the question is absolutely correct. I have triple checked with the source. Anyhow any clue even in the D notation would also be helpful. I will also be glad if you give me the equivalent D notation equation. – 2017-02-28
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0If you have a $y$ inside the brackets, rather than just the one $y$ at the end, the equation becomes non-linear. What I am thinking is that the brackets with derivatives represent a differential operator acting on a single $y$ at the end. If so, your equation is a linear differential equation of the sixth degree, and could be solved (amongst other methods) by solving three second degree equations of the kind you already know how to deal with. e.g. set $z=(\frac {d^2y}{dx^2}+1)$ etc – 2017-02-28
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0@kloikfriend: as written, this ODE is absolutely untractable. Where is that monster coming from ? [I keep thinking that you copied the wrong way.] – 2017-02-28
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0Thanks @MarkBennet . "your equation is a linear differential equation of the fourth degree" caught my eye. I 'll try your suggestion. Thanks. – 2017-02-28
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0@YvesDaoust haha I like it when you call it monster, check this link — [link](https://webservices.ignou.ac.in/assignments/bsc/2017/bsc/MATHS/Maths%202017%20(E)/MTE-08-ENGLISH.pdf) Question 6) a) i) Another monster is also lying below it. – 2017-02-28
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0Waw, that's a high level exam. I confirm that I have absolutely no clue how to attack the beast. (And I keep wondering about typos in the problem statement; actually you don't expect the first two $y$'s to be there.) – 2017-02-28
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0Thanks all. Although I will try to solve it, the question is still open for who want to post a solution. – 2017-02-28
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0The equation gets to sixth degree (I missed the second square), but can be solved by applying insights and methods developed on second degree equations. – 2017-02-28
2 Answers
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Hint:
Assuming the sixth order linear ODE
$$\left({d\over dx}-1\right)^2\left({d^2 \over dx^2} +1\right)^2y = \frac{1-\cos x}2+e^x+x$$ where the characteristic polynomial has the double roots $1$ and $\pm i$, we can infer a general solution that includes the following terms:
$$1,x,e^x,xe^x,x^2e^x,\cos x,\sin x,x\cos x,x\sin x,x^2\cos x,x^2\sin x.$$
You can solve by the method of undeterminate coefficients.
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Hint (not a complete answer just an observation):
This ODE is not trivial at all. By inspection or by solving $y''+1=0$, you can guess one solution of the homogenous ODE.
$$y_h=-\frac{1}{2}x^2+bx+c$$
which eliminates the second bracket.