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calculus help!!!

A

alphatarget1

Diamond Member
yeah it's pretty pathetic, 1st semester calculus 😛

the problem is as follows: the equation y''+y'-2y=x^2 is called a differential equation because it involves an unknow function y and its derivatives y' and y". Find constants A, B and C such that the function y=Ax^2+Bx+C satisfies this equation.

here's the deal, i'm on chapter 3 and they aren't going to teach differential equations until chapter 7. I was thinking of putting y by itself and then get its derivatives (y' & y") but I don't know how that'd relate to the latter equation (y=Ax^2+Bx+C). We just learned power rule ,quotient rule and exponent rule and I have no clue how to do this

any help is appreciated 🙂
 
they don't, i'm on chapter 3 and they stated that they'll talk about differential equations in chapter 7.

I have no idea how y''+y'-2y=x^2 and y=Ax^2+Bx+C are related :/
 
It's not a differential equation problem.

y'' + y' - 2y = x^2
y = Ax^2 + Bx + C
y' = 2Ax + B
y" = 2A

Therefore,
2A + 2Ax + B - 2Ax^2 - 2Bx - 2C = x^2

-2A = 1
-2B + 2A = 0
2A - 2C + B = 0

EDIT3: I'll just let you solve that on your own, cause obviously my brain isn't working and NyQuil is kicking in. The answer should be obvious from here.
EDIT: Fixed math error.
EDIT2: HOLD! Another math error.
 
yeah i was working on it on my own i got the following

A= -1/2
B= -1/2
C= -3/4

i didn't notice the 2nd equation, i got the derivative of Ax^2+Bx+C which is 2Ax+B and then the derivative of that which is 2A

and plugged into the equation
2A+2Ax+B-2(Ax^2+Bx+C)=x^2
2A+2Ax+B-2Ax^2-2Bx-2C=x^2
2(-1/2)+2(-1x/2)-1/2-2(-1x^2/2)-2(-1x/2)-3/2=x^2 which gives me
-1-x-(1/2)+x^2+x-(3/2)=x^2

they all cancel out, leaving x^2=x^2

is that right?
 
Originally posted by: kenleung
yeah i was working on it on my own i got the following

A= -1/2
B= -1/2
C= -3/4

i didn't notice the 2nd equation, i got the derivative of Ax^2+Bx+C which is 2Ax+B and then the derivative of that which is 2A

and plugged into the equation
2A+2Ax+B-2(Ax^2+Bx+C)=x^2
2A+2Ax+B-2Ax^2-2Bx-2C=x^2
2(-1/2)+2(-1x/2)-1/2-2(-1x^2/2)-2(-1x/2)-3/2=x^2 which gives me
-1-x-(1/2)+x^2+x-(3/2)=x^2

they all cancel out, leaving x^2=x^2

is that right?
Click to expand...

Yes.

 
Originally posted by: Rkonster
It's not a differential equation problem.

y'' + y' - 2y = x^2
y = Ax^2 + Bx + C
y' = 2Ax + B
y" = 2A

Therefore,
2A + 2Ax + B - 2Ax^2 - 2Bx - 2C = x^2

-2A = 1
-2B + 2A = 0
2A - 2C + B = 0

EDIT3: I'll just let you solve that on your own, cause obviously my brain isn't working and NyQuil is kicking in. The answer should be obvious from here.
EDIT: Fixed math error.
EDIT2: HOLD! Another math error.
Click to expand...

Its not a DE problem? I'm taking a DE class now, and IIRC its called the superposition approach for undetermined coefficients. Sure they give you the general form of the solution (so you don't have to "guess"), but its still definately a DE.

 
Originally posted by: wviperw
Originally posted by: Rkonster
It's not a differential equation problem.

y'' + y' - 2y = x^2
y = Ax^2 + Bx + C
y' = 2Ax + B
y" = 2A

Therefore,
2A + 2Ax + B - 2Ax^2 - 2Bx - 2C = x^2

-2A = 1
-2B + 2A = 0
2A - 2C + B = 0

EDIT3: I'll just let you solve that on your own, cause obviously my brain isn't working and NyQuil is kicking in. The answer should be obvious from here.
EDIT: Fixed math error.
EDIT2: HOLD! Another math error.
Click to expand...

Its not a DE problem? I'm taking a DE class now, and IIRC its called the superposition approach for undetermined coefficients. Sure they give you the general form of the solution (so you don't have to "guess"), but its still definately a DE.
Click to expand...


Yeah, this is a non-homogeneous differential equation, which can be solved using the method of undeterrmined coefficients or variation of paramters. All I meant was that for the purposes of Calc I, you do not need to concern yourself with it.
 
Originally posted by: Rkonster
Originally posted by: wviperw
Originally posted by: Rkonster
It's not a differential equation problem.

y'' + y' - 2y = x^2
y = Ax^2 + Bx + C
y' = 2Ax + B
y" = 2A

Therefore,
2A + 2Ax + B - 2Ax^2 - 2Bx - 2C = x^2

-2A = 1
-2B + 2A = 0
2A - 2C + B = 0

EDIT3: I'll just let you solve that on your own, cause obviously my brain isn't working and NyQuil is kicking in. The answer should be obvious from here.
EDIT: Fixed math error.
EDIT2: HOLD! Another math error.
Click to expand...

Its not a DE problem? I'm taking a DE class now, and IIRC its called the superposition approach for undetermined coefficients. Sure they give you the general form of the solution (so you don't have to "guess"), but its still definately a DE.
Click to expand...


Yeah, this is a non-homogeneous differential equation, which can be solved using the method of undeterrmined coefficients or variation of paramters. All I meant was that for the purposes of Calc I, you do not need to concern yourself with it.
Click to expand...


Basically.
Use your characteriistic equation to find a fundamental set of solutions, for the corresponding homogenious equation, y''+y'-2 =0. Thus r^2+r-2 = 0. Thus x is 1 or -2 and your fundamental set of solution is e^-t and e^2t. then solve for the particulay solution phi(x) = x^2. Using the method of undetermined coefficents for a second-degree polynomial, y=Ax^2+Bx+C, y'=2Ax+B, y''=2A, thus (2A + 2Ax+B - 2Ax^2-2Bx-2C. Collecting terms, x^2(-2A) + x(2A-2B)+(2A+B-2C) = x^2.

A = -1/2;
2A-2B = 0; B = -1/2;
2A+B-2C = 0, C = -3/4

Thus, adding Yparticular and Yhomogeneous, the solution is c1e^2t+c2e^-t-(1/2)x^2-(1/2)x-3/4.
 
Originally posted by: wviperw
Originally posted by: Rkonster
It's not a differential equation problem.

y'' + y' - 2y = x^2
y = Ax^2 + Bx + C
y' = 2Ax + B
y" = 2A

Therefore,
2A + 2Ax + B - 2Ax^2 - 2Bx - 2C = x^2

-2A = 1
-2B + 2A = 0
2A - 2C + B = 0

EDIT3: I'll just let you solve that on your own, cause obviously my brain isn't working and NyQuil is kicking in. The answer should be obvious from here.
EDIT: Fixed math error.
EDIT2: HOLD! Another math error.
Click to expand...

Its not a DE problem? I'm taking a DE class now, and IIRC its called the superposition approach for undetermined coefficients. Sure they give you the general form of the solution (so you don't have to "guess"), but its still definately a DE.
Click to expand...


Of course its a DE. But notice they did not ask him to find the complete solution. In fact, even when he finds A, B, and C, he hasn't found the complete solution. (You also have to solve the homogenous problem y'' + y' - 2y = 0 to find the complete solution).

This is a GOOD question for someone learning to take derivatives. The ENTIRE solution is just "take the derivatives of the given function that are needed, plug it into the equation, and solve for the unknown coefficients).
 
Originally posted by: Turin39789
where are you taking this at? Everywhere I have seen there is a class called calc 4 aka Differential equations
Click to expand...

calc 4 = vector calculus
diff eq = something separate
 
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