Finding Maximal solution of Y

[Superfastjellyfish]

Questions 31 - 33: Consider the differential equation:

. . . . .\(\displaystyle y"\, -\, 3y'\, +\, 2y\, =\, \sin(x)\)

...with initial conditions \(\displaystyle \, y(0)\, =\, 1\, \) and \(\displaystyle \, y'(0)\, =\, 2.\,\) Find the maximal solution \(\displaystyle \,y\,\) of this problem and answer the following questions.

31. The value of \(\displaystyle \, 10\, y(\pi)\,\) is equal to:

a) \(\displaystyle 10\, +\, e\) . . .b) \(\displaystyle e^{\pi}\, \left(12e^{\pi}\, -\, 5\right)\, -\, 3\) . . .c) \(\displaystyle 3\, -\, 10\, \sin(\pi)\, +\, e^{2\pi}\) . . .d) \(\displaystyle \pi e^{-2 \pi}\, -\, 3\) . . .e) \(\displaystyle \dfrac{3}{4}\, -\, \dfrac{e^{\pi}}{4}\)

32. The solution \(\displaystyle \, y\, \) is on its domain:

a) decreasing . . .b) constant . . .c) \(\displaystyle 2 \pi\)-periodic . . .d) increasing . . .e) none of these

33. The limit \(\displaystyle \displaystyle{ \lim_{x\, \rightarrow \, +\infty} \,}\) \(\displaystyle \dfrac{y(x)}{e^{2x}}\,\) is equal to:

a) \(\displaystyle +\infty\) . . .b) \(\displaystyle \dfrac{3}{2}\) . . .c) \(\displaystyle 0\) . . .d) \(\displaystyle \dfrac{6}{5}\) . . .e) does not exist

Ive learned this stuff 7 years ago, if somebody can show me how it is done Id be so grateful !

 

[stapel]

Questions 31 - 33: Consider the differential equation \(\displaystyle \,y''\, -\, 3y'\, +\, 2y\, =\, \sin(x)\,\) with initial conditions \(\displaystyle \, y(0)\, =\, 1\, \) and \(\displaystyle \, y'(0)\, =\, 2.\,\) Find the maximal solution \(\displaystyle \,y\,\) of this problem and answer the following questions.

31. The value of \(\displaystyle \, 10\, y(\pi)\,\) is equal to:
a) \(\displaystyle 10\, +\, e\) . . .b) \(\displaystyle e^{\pi}\, \left(12e^{\pi}\, -\, 5\right)\, -\, 3\) . . .c) \(\displaystyle 3\, -\, 10\, \sin(\pi)\, +\, e^{2\pi}\) . . .d) \(\displaystyle \pi e^{-2 \pi}\, -\, 3\) . . .e) \(\displaystyle \dfrac{3}{4}\, -\, \dfrac{e^{\pi}}{4}\)

32. The solution \(\displaystyle \, y\, \) is on its domain:
a) decreasing . . .b) constant . . .c) \(\displaystyle 2 \pi\)-periodic . . .d) increasing . . .e) none of these

33. The limit \(\displaystyle \displaystyle{ \lim_{x\, \rightarrow \, +\infty} \,}\) \(\displaystyle \dfrac{y(x)}{e^{2x}}\,\) is equal to:
a) \(\displaystyle +\infty\) . . .b) \(\displaystyle \dfrac{3}{2}\) . . .c) \(\displaystyle 0\) . . .d) \(\displaystyle \dfrac{6}{5}\) . . .e) does not exist

Ive learned this stuff 7 years ago, if somebody can show me how it is done Id be so grateful !

Are you saying that you don't remember enough even to attempt a start at any of this, and are needing lessons to re-learn this topic? Thank you!

 

[HallsofIvy]

If you 'learned this stuff seven years ago", why do you need to solve such a problem now? If you are taking a new class, then its textbook should give you ideas on solving it.

In any case, this is a "second order ordinary, linear, differential equation with constant coefficients". It is "non-homogeneous" because there is a term that does not involve y, sin(x). The simplest way to solve it is to first solve the corresponding homogeneous" equation, \(\displaystyle y''- 3y'+ 2y= 0\). If you were to try a solution of the form \(\displaystyle y= e^{rx}\) then you would see that you get its "characteristic equation", \(\displaystyle (r^2- 3r+ 2= 0\), a quadratic equation with two roots. Calling those two roots \(\displaystyle r_1\) and \(\displaystyle r_2\), then the solution to the corresponding homogeneous equation is \(\displaystyle C_1e^{r_1x}+ C_2e^{r_2x}\) where \(\displaystyle C_1\) and \(\displaystyle C_2\) can be any constants.

To find the general solution to the entire equation, we need only add to that a single function that satisfies the entire equation. Try something of the form \(\displaystyle y(x)= A cos(x)+ B sin(x)\).

 

[Superfastjellyfish]

In any case, this is a "second order ordinary, linear, differential equation with constant coefficients". It is "non-homogeneous" because there is a term that does not involve y, sin(x). The simplest way to solve it is to first solve the corresponding homogeneous" equation, y″−3y′+2y=0. If you were to try a solution of the form y=erx then you would see that you get its "characteristic equation", (r2−3r+2=0, a quadratic equation with two roots. Calling those two roots r1 and r2, then the solution to the corresponding homogeneous equation is C1er1x+C2er2x where C1 and C2 can be any constants.

To find the general solution to the entire equation, we need only add to that a single function that satisfies the entire equation. Try something of the form y(x)=Acos(x)+Bsin(x).

Thank you everyone.
The fact is that books are just theory and u need practice to remember, Ive not used it since Ive learned it and now when I see the way how I should start and how it must go, I remember some stuff and I can try myself.