$y'' + y' -2y = x^2$, find $A, B$, & and $C$ such that $y = Ax^2+Bx+C$ satisfies this equation. The 2019 Stack Overflow Developer Survey Results Are In Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Show that Bessel function $J_n(x)$ satisfies Bessel's differential equation.Find tangent line at given points, no function equationFokker-Planck equation - find probability density functionFind the value of $n$ such that the Maclaurin polynomial error is within a boundGet the general solution for this differential equationA differentiable function satisfies given conditions. Find the approximated valuesShow that this function satisfies the Laplace EquationShow that a given function satisfies a given equationFind Polar Differential Equation that satisfies Bifurcation DiagramThe differentiable equation satisfies the equation…
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$y'' + y' -2y = x^2$, find $A, B$, & and $C$ such that $y = Ax^2+Bx+C$ satisfies this equation.
The 2019 Stack Overflow Developer Survey Results Are In
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Show that Bessel function $J_n(x)$ satisfies Bessel's differential equation.Find tangent line at given points, no function equationFokker-Planck equation - find probability density functionFind the value of $n$ such that the Maclaurin polynomial error is within a boundGet the general solution for this differential equationA differentiable function satisfies given conditions. Find the approximated valuesShow that this function satisfies the Laplace EquationShow that a given function satisfies a given equationFind Polar Differential Equation that satisfies Bifurcation DiagramThe differentiable equation satisfies the equation…
$begingroup$
I am doing an extra credit problem for college. I don't expect anyone to solve it for me, but I would really appreciate being given some hints.
The problem:
$y'' + y' -2y = x^2$, find $A, B$, & and $C$ such that the function $y = Ax^2+Bx+C$ satisfies this equation.
I understand how to find derivatives if I know the function, but this is stumping me.
calculus ordinary-differential-equations derivatives
asked Mar 24 at 19:25
LuminousNutriaLuminousNutria
55712
$endgroup$
add a comment |
$begingroup$
I am doing an extra credit problem for college. I don't expect anyone to solve it for me, but I would really appreciate being given some hints.
The problem:
$y'' + y' -2y = x^2$, find $A, B$, & and $C$ such that the function $y = Ax^2+Bx+C$ satisfies this equation.
I understand how to find derivatives if I know the function, but this is stumping me.
calculus ordinary-differential-equations derivatives
asked Mar 24 at 19:25
LuminousNutriaLuminousNutria
55712
$endgroup$
add a comment |
$begingroup$
I am doing an extra credit problem for college. I don't expect anyone to solve it for me, but I would really appreciate being given some hints.
The problem:
$y'' + y' -2y = x^2$, find $A, B$, & and $C$ such that the function $y = Ax^2+Bx+C$ satisfies this equation.
I understand how to find derivatives if I know the function, but this is stumping me.
calculus ordinary-differential-equations derivatives
asked Mar 24 at 19:25
LuminousNutriaLuminousNutria
55712
$endgroup$
I am doing an extra credit problem for college. I don't expect anyone to solve it for me, but I would really appreciate being given some hints.
The problem:
$y'' + y' -2y = x^2$, find $A, B$, & and $C$ such that the function $y = Ax^2+Bx+C$ satisfies this equation.
I understand how to find derivatives if I know the function, but this is stumping me.
calculus ordinary-differential-equations derivatives
calculus ordinary-differential-equations derivatives
asked Mar 24 at 19:25
LuminousNutriaLuminousNutria
55712
asked Mar 24 at 19:25
LuminousNutriaLuminousNutria
55712
asked Mar 24 at 19:25
LuminousNutriaLuminousNutria
55712
asked Mar 24 at 19:25
LuminousNutriaLuminousNutria
55712
asked Mar 24 at 19:25
LuminousNutriaLuminousNutria
55712
55712
add a comment |
add a comment |
4 Answers
4
active
oldest
votes
$begingroup$
Assuming that $y = y(x)$ is of the given form, we need only substitute $y$ into the differential equation
$$
y^primeprime + y^prime - 2y = x^2
$$
and solve for $A,B$ and $C$. Now, we have
beginalignlabeleq:1tag1
y^prime(x) = 2Ax + B quad textand quad y^primeprime(x) = 2A.
endalign
So, if $y$ solves the differential equation, then
beginalign
x^2 = y^primeprime + y^prime - 2y &= 2A + (2Ax+B) -2left(Ax^2 + Bx + C right)labeleq:2tag2\
&= -2Ax^2 + (2A-2B)x + left( 2A + B -2Cright)labeleq:3tag3.
endalign
Comparing the coefficients of this equation, we find that
beginalign*
begincases
-2A = 1,\
2A-2B = 0,\
2A+B-2C = 0.
endcases
endalign*
Solving this system will give the explicit values of $A,B$ and $C$ required.
answered Mar 24 at 19:29
rolandcyprolandcyp
1
$endgroup$
$begingroup$
What do you replace $y^primeprime + y^prime - 2y = x^2$ with, when you solve for $A, B$, and $C$?
$endgroup$
– LuminousNutria
Mar 24 at 19:40
$begingroup$
I substitute in the expressions for $y,y^prime$ and $y^primeprime$ in terms of $x$, respectively.
$endgroup$
– rolandcyp
Mar 24 at 19:42
$begingroup$
Oh, so you do $y''(x) + y'(x) - 2y(x) = x^2$ then get rid of $x^2$ and solve for $y'(x)$ and $y''(x)$ respectively?
$endgroup$
– LuminousNutria
Mar 24 at 19:43
$begingroup$
Not exactly. I'm not getting rid of anything, I am only comparing coefficients. If two polynomials are everywhere equal, then their coefficients must agree. Using this, I'm solving for $A,B$ and $C$.
$endgroup$
– rolandcyp
Mar 24 at 19:44
1
$begingroup$
ohh! I get it! thank you!
$endgroup$
– LuminousNutria
Mar 24 at 19:56
|
show 7 more comments
$begingroup$
If you plug the quadratic polynomial into your differential equation you will get
$$2A+(2Ax+B)-2(Ax^2+Bx+C)=x^2.$$
Compare the coefficients and determine $A$, $B$, $C$.
answered Mar 24 at 19:29
MachineLearnerMachineLearner
1,386212
$endgroup$
add a comment |
$begingroup$
With
$y = Ax^2 + Bx + C, tag 1$
we may substitute $y$, $y'$, and $y''$ into
$y'' + y' - 2y = x^2, tag 2$
viz,
$y' = 2Ax + B, tag 3$
$y'' = 2A, tag 4$
$y'' + y' - 2y = 2A + (2Ax + B) - 2(Ax^2 + Bx + C) = x^2; tag 5$
we group together like powers of $x$:
$-2Ax^2 + 2(A - B) x + (2A + B -2C) = x^2, tag 5$
from which we infer
$-2A = 1, tag 6$
$2(A - B) = 0, tag 7$
$2A + B - 2C = 0; tag 8$
thus
$A = B = -dfrac12, tag 9$
$C = -dfrac34, tag10$
and of course
$y = -dfrac12x^2 - dfrac12x - dfrac34. tag11$
We Check:
From (11),
$y' = -x - dfrac12, tag12$
$y'' = -1, tag13$
$y'' + y' - 2y$
$= -1 - x - dfrac12 - 2(-dfrac12x^2 - dfrac12x - dfrac34) = -1 - x - dfrac12 + x^2 + x + dfrac32 = x^2. tag14$
answered Mar 24 at 20:21
Robert LewisRobert Lewis
48.9k23168
$endgroup$
1
$begingroup$
How did you do steps 6, 7, and 8?
$endgroup$
– LuminousNutria
Mar 24 at 20:35
1
$begingroup$
@LuminousNutria: I simply took (5) and equated the coefficients of powers of $x$--$x^0 = 1, ; x^1, ; x^2$--occurring on each side.
$endgroup$
– Robert Lewis
Mar 24 at 20:40
$begingroup$
So, $-2A=1$ because the coefficient of $x^2$ is 1? And $2(A - B) = 0$ because the coefficients of $2x - 2x$ equal $1 - 1$ and are therefore zero? And $(2A+B-2C) = 0$ because it has no powers of x?
$endgroup$
– LuminousNutria
Mar 24 at 20:55
1
$begingroup$
@LuminousNutria: yes! Basically what we're doing is using the fact that each side is a quadratic polynomial in $x$, since they are equal, the coefficients have to be the same. Equal polynomials have equal coefficients! Cheers!
$endgroup$
– Robert Lewis
Mar 24 at 21:05
add a comment |
$begingroup$
Consider $z^2+z-2=(z+2)(z-1)=-2(1+frac12z)(1-z)$. Its inverse (as a formal power series) is
$$
-frac12(1+z+z^2+dotsb)left(1-frac12z+frac14z^2+dotsbright)
=-dfrac12-dfrac14z-dfrac38z^2+dotsb
$$
Interpret $z$ as the differentiation operator and “multiply” by $x^2$ (that's why we can discard higher order terms):
$$
-dfrac12x^2-dfrac142x-dfrac382=-dfrac12x^2-dfrac12x-dfrac34
$$
answered Mar 24 at 20:41
egregegreg
186k1486208
$endgroup$
add a comment |
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Assuming that $y = y(x)$ is of the given form, we need only substitute $y$ into the differential equation
$$
y^primeprime + y^prime - 2y = x^2
$$
and solve for $A,B$ and $C$. Now, we have
beginalignlabeleq:1tag1
y^prime(x) = 2Ax + B quad textand quad y^primeprime(x) = 2A.
endalign
So, if $y$ solves the differential equation, then
beginalign
x^2 = y^primeprime + y^prime - 2y &= 2A + (2Ax+B) -2left(Ax^2 + Bx + C right)labeleq:2tag2\
&= -2Ax^2 + (2A-2B)x + left( 2A + B -2Cright)labeleq:3tag3.
endalign
Comparing the coefficients of this equation, we find that
beginalign*
begincases
-2A = 1,\
2A-2B = 0,\
2A+B-2C = 0.
endcases
endalign*
Solving this system will give the explicit values of $A,B$ and $C$ required.
answered Mar 24 at 19:29
rolandcyprolandcyp
1
$endgroup$
$begingroup$
What do you replace $y^primeprime + y^prime - 2y = x^2$ with, when you solve for $A, B$, and $C$?
$endgroup$
– LuminousNutria
Mar 24 at 19:40
$begingroup$
I substitute in the expressions for $y,y^prime$ and $y^primeprime$ in terms of $x$, respectively.
$endgroup$
– rolandcyp
Mar 24 at 19:42
$begingroup$
Oh, so you do $y''(x) + y'(x) - 2y(x) = x^2$ then get rid of $x^2$ and solve for $y'(x)$ and $y''(x)$ respectively?
$endgroup$
– LuminousNutria
Mar 24 at 19:43
$begingroup$
Not exactly. I'm not getting rid of anything, I am only comparing coefficients. If two polynomials are everywhere equal, then their coefficients must agree. Using this, I'm solving for $A,B$ and $C$.
$endgroup$
– rolandcyp
Mar 24 at 19:44
1
$begingroup$
ohh! I get it! thank you!
$endgroup$
– LuminousNutria
Mar 24 at 19:56
|
show 7 more comments
$begingroup$
Assuming that $y = y(x)$ is of the given form, we need only substitute $y$ into the differential equation
$$
y^primeprime + y^prime - 2y = x^2
$$
and solve for $A,B$ and $C$. Now, we have
beginalignlabeleq:1tag1
y^prime(x) = 2Ax + B quad textand quad y^primeprime(x) = 2A.
endalign
So, if $y$ solves the differential equation, then
beginalign
x^2 = y^primeprime + y^prime - 2y &= 2A + (2Ax+B) -2left(Ax^2 + Bx + C right)labeleq:2tag2\
&= -2Ax^2 + (2A-2B)x + left( 2A + B -2Cright)labeleq:3tag3.
endalign
Comparing the coefficients of this equation, we find that
beginalign*
begincases
-2A = 1,\
2A-2B = 0,\
2A+B-2C = 0.
endcases
endalign*
Solving this system will give the explicit values of $A,B$ and $C$ required.
answered Mar 24 at 19:29
rolandcyprolandcyp
1
$endgroup$
$begingroup$
What do you replace $y^primeprime + y^prime - 2y = x^2$ with, when you solve for $A, B$, and $C$?
$endgroup$
– LuminousNutria
Mar 24 at 19:40
$begingroup$
I substitute in the expressions for $y,y^prime$ and $y^primeprime$ in terms of $x$, respectively.
$endgroup$
– rolandcyp
Mar 24 at 19:42
$begingroup$
Oh, so you do $y''(x) + y'(x) - 2y(x) = x^2$ then get rid of $x^2$ and solve for $y'(x)$ and $y''(x)$ respectively?
$endgroup$
– LuminousNutria
Mar 24 at 19:43
$begingroup$
Not exactly. I'm not getting rid of anything, I am only comparing coefficients. If two polynomials are everywhere equal, then their coefficients must agree. Using this, I'm solving for $A,B$ and $C$.
$endgroup$
– rolandcyp
Mar 24 at 19:44
1
$begingroup$
ohh! I get it! thank you!
$endgroup$
– LuminousNutria
Mar 24 at 19:56
|
show 7 more comments
$begingroup$
Assuming that $y = y(x)$ is of the given form, we need only substitute $y$ into the differential equation
$$
y^primeprime + y^prime - 2y = x^2
$$
and solve for $A,B$ and $C$. Now, we have
beginalignlabeleq:1tag1
y^prime(x) = 2Ax + B quad textand quad y^primeprime(x) = 2A.
endalign
So, if $y$ solves the differential equation, then
beginalign
x^2 = y^primeprime + y^prime - 2y &= 2A + (2Ax+B) -2left(Ax^2 + Bx + C right)labeleq:2tag2\
&= -2Ax^2 + (2A-2B)x + left( 2A + B -2Cright)labeleq:3tag3.
endalign
Comparing the coefficients of this equation, we find that
beginalign*
begincases
-2A = 1,\
2A-2B = 0,\
2A+B-2C = 0.
endcases
endalign*
Solving this system will give the explicit values of $A,B$ and $C$ required.
answered Mar 24 at 19:29
rolandcyprolandcyp
1
$endgroup$
Assuming that $y = y(x)$ is of the given form, we need only substitute $y$ into the differential equation
$$
y^primeprime + y^prime - 2y = x^2
$$
and solve for $A,B$ and $C$. Now, we have
beginalignlabeleq:1tag1
y^prime(x) = 2Ax + B quad textand quad y^primeprime(x) = 2A.
endalign
So, if $y$ solves the differential equation, then
beginalign
x^2 = y^primeprime + y^prime - 2y &= 2A + (2Ax+B) -2left(Ax^2 + Bx + C right)labeleq:2tag2\
&= -2Ax^2 + (2A-2B)x + left( 2A + B -2Cright)labeleq:3tag3.
endalign
Comparing the coefficients of this equation, we find that
beginalign*
begincases
-2A = 1,\
2A-2B = 0,\
2A+B-2C = 0.
endcases
endalign*
Solving this system will give the explicit values of $A,B$ and $C$ required.
answered Mar 24 at 19:29
rolandcyprolandcyp
1
edited Mar 24 at 19:49
answered Mar 24 at 19:29
rolandcyprolandcyp
1
answered Mar 24 at 19:29
rolandcyprolandcyp
1
answered Mar 24 at 19:29
rolandcyprolandcyp
1
1
$begingroup$
What do you replace $y^primeprime + y^prime - 2y = x^2$ with, when you solve for $A, B$, and $C$?
$endgroup$
– LuminousNutria
Mar 24 at 19:40
$begingroup$
I substitute in the expressions for $y,y^prime$ and $y^primeprime$ in terms of $x$, respectively.
$endgroup$
– rolandcyp
Mar 24 at 19:42
$begingroup$
Oh, so you do $y''(x) + y'(x) - 2y(x) = x^2$ then get rid of $x^2$ and solve for $y'(x)$ and $y''(x)$ respectively?
$endgroup$
– LuminousNutria
Mar 24 at 19:43
$begingroup$
Not exactly. I'm not getting rid of anything, I am only comparing coefficients. If two polynomials are everywhere equal, then their coefficients must agree. Using this, I'm solving for $A,B$ and $C$.
$endgroup$
– rolandcyp
Mar 24 at 19:44
1
$begingroup$
ohh! I get it! thank you!
$endgroup$
– LuminousNutria
Mar 24 at 19:56
|
show 7 more comments
$begingroup$
What do you replace $y^primeprime + y^prime - 2y = x^2$ with, when you solve for $A, B$, and $C$?
$endgroup$
– LuminousNutria
Mar 24 at 19:40
$begingroup$
I substitute in the expressions for $y,y^prime$ and $y^primeprime$ in terms of $x$, respectively.
$endgroup$
– rolandcyp
Mar 24 at 19:42
$begingroup$
Oh, so you do $y''(x) + y'(x) - 2y(x) = x^2$ then get rid of $x^2$ and solve for $y'(x)$ and $y''(x)$ respectively?
$endgroup$
– LuminousNutria
Mar 24 at 19:43
$begingroup$
Not exactly. I'm not getting rid of anything, I am only comparing coefficients. If two polynomials are everywhere equal, then their coefficients must agree. Using this, I'm solving for $A,B$ and $C$.
$endgroup$
– rolandcyp
Mar 24 at 19:44
1
$begingroup$
ohh! I get it! thank you!
$endgroup$
– LuminousNutria
Mar 24 at 19:56
$begingroup$
What do you replace $y^primeprime + y^prime - 2y = x^2$ with, when you solve for $A, B$, and $C$?
$endgroup$
– LuminousNutria
Mar 24 at 19:40
$begingroup$
What do you replace $y^primeprime + y^prime - 2y = x^2$ with, when you solve for $A, B$, and $C$?
$endgroup$
– LuminousNutria
Mar 24 at 19:40
$begingroup$
I substitute in the expressions for $y,y^prime$ and $y^primeprime$ in terms of $x$, respectively.
$endgroup$
– rolandcyp
Mar 24 at 19:42
$begingroup$
I substitute in the expressions for $y,y^prime$ and $y^primeprime$ in terms of $x$, respectively.
$endgroup$
– rolandcyp
Mar 24 at 19:42
$begingroup$
Oh, so you do $y''(x) + y'(x) - 2y(x) = x^2$ then get rid of $x^2$ and solve for $y'(x)$ and $y''(x)$ respectively?
$endgroup$
– LuminousNutria
Mar 24 at 19:43
$begingroup$
Oh, so you do $y''(x) + y'(x) - 2y(x) = x^2$ then get rid of $x^2$ and solve for $y'(x)$ and $y''(x)$ respectively?
$endgroup$
– LuminousNutria
Mar 24 at 19:43
$begingroup$
Not exactly. I'm not getting rid of anything, I am only comparing coefficients. If two polynomials are everywhere equal, then their coefficients must agree. Using this, I'm solving for $A,B$ and $C$.
$endgroup$
– rolandcyp
Mar 24 at 19:44
$begingroup$
Not exactly. I'm not getting rid of anything, I am only comparing coefficients. If two polynomials are everywhere equal, then their coefficients must agree. Using this, I'm solving for $A,B$ and $C$.
$endgroup$
– rolandcyp
Mar 24 at 19:44
1
1
$begingroup$
ohh! I get it! thank you!
$endgroup$
– LuminousNutria
Mar 24 at 19:56
$begingroup$
ohh! I get it! thank you!
$endgroup$
– LuminousNutria
Mar 24 at 19:56
|
show 7 more comments
$begingroup$
If you plug the quadratic polynomial into your differential equation you will get
$$2A+(2Ax+B)-2(Ax^2+Bx+C)=x^2.$$
Compare the coefficients and determine $A$, $B$, $C$.
answered Mar 24 at 19:29
MachineLearnerMachineLearner
1,386212
$endgroup$
add a comment |
$begingroup$
If you plug the quadratic polynomial into your differential equation you will get
$$2A+(2Ax+B)-2(Ax^2+Bx+C)=x^2.$$
Compare the coefficients and determine $A$, $B$, $C$.
answered Mar 24 at 19:29
MachineLearnerMachineLearner
1,386212
$endgroup$
add a comment |
$begingroup$
If you plug the quadratic polynomial into your differential equation you will get
$$2A+(2Ax+B)-2(Ax^2+Bx+C)=x^2.$$
Compare the coefficients and determine $A$, $B$, $C$.
answered Mar 24 at 19:29
MachineLearnerMachineLearner
1,386212
$endgroup$
If you plug the quadratic polynomial into your differential equation you will get
$$2A+(2Ax+B)-2(Ax^2+Bx+C)=x^2.$$
Compare the coefficients and determine $A$, $B$, $C$.
answered Mar 24 at 19:29
MachineLearnerMachineLearner
1,386212
answered Mar 24 at 19:29
MachineLearnerMachineLearner
1,386212
answered Mar 24 at 19:29
MachineLearnerMachineLearner
1,386212
answered Mar 24 at 19:29
MachineLearnerMachineLearner
1,386212
1,386212
add a comment |
add a comment |
$begingroup$
With
$y = Ax^2 + Bx + C, tag 1$
we may substitute $y$, $y'$, and $y''$ into
$y'' + y' - 2y = x^2, tag 2$
viz,
$y' = 2Ax + B, tag 3$
$y'' = 2A, tag 4$
$y'' + y' - 2y = 2A + (2Ax + B) - 2(Ax^2 + Bx + C) = x^2; tag 5$
we group together like powers of $x$:
$-2Ax^2 + 2(A - B) x + (2A + B -2C) = x^2, tag 5$
from which we infer
$-2A = 1, tag 6$
$2(A - B) = 0, tag 7$
$2A + B - 2C = 0; tag 8$
thus
$A = B = -dfrac12, tag 9$
$C = -dfrac34, tag10$
and of course
$y = -dfrac12x^2 - dfrac12x - dfrac34. tag11$
We Check:
From (11),
$y' = -x - dfrac12, tag12$
$y'' = -1, tag13$
$y'' + y' - 2y$
$= -1 - x - dfrac12 - 2(-dfrac12x^2 - dfrac12x - dfrac34) = -1 - x - dfrac12 + x^2 + x + dfrac32 = x^2. tag14$
answered Mar 24 at 20:21
Robert LewisRobert Lewis
48.9k23168
$endgroup$
1
$begingroup$
How did you do steps 6, 7, and 8?
$endgroup$
– LuminousNutria
Mar 24 at 20:35
1
$begingroup$
@LuminousNutria: I simply took (5) and equated the coefficients of powers of $x$--$x^0 = 1, ; x^1, ; x^2$--occurring on each side.
$endgroup$
– Robert Lewis
Mar 24 at 20:40
$begingroup$
So, $-2A=1$ because the coefficient of $x^2$ is 1? And $2(A - B) = 0$ because the coefficients of $2x - 2x$ equal $1 - 1$ and are therefore zero? And $(2A+B-2C) = 0$ because it has no powers of x?
$endgroup$
– LuminousNutria
Mar 24 at 20:55
1
$begingroup$
@LuminousNutria: yes! Basically what we're doing is using the fact that each side is a quadratic polynomial in $x$, since they are equal, the coefficients have to be the same. Equal polynomials have equal coefficients! Cheers!
$endgroup$
– Robert Lewis
Mar 24 at 21:05
add a comment |
$begingroup$
With
$y = Ax^2 + Bx + C, tag 1$
we may substitute $y$, $y'$, and $y''$ into
$y'' + y' - 2y = x^2, tag 2$
viz,
$y' = 2Ax + B, tag 3$
$y'' = 2A, tag 4$
$y'' + y' - 2y = 2A + (2Ax + B) - 2(Ax^2 + Bx + C) = x^2; tag 5$
we group together like powers of $x$:
$-2Ax^2 + 2(A - B) x + (2A + B -2C) = x^2, tag 5$
from which we infer
$-2A = 1, tag 6$
$2(A - B) = 0, tag 7$
$2A + B - 2C = 0; tag 8$
thus
$A = B = -dfrac12, tag 9$
$C = -dfrac34, tag10$
and of course
$y = -dfrac12x^2 - dfrac12x - dfrac34. tag11$
We Check:
From (11),
$y' = -x - dfrac12, tag12$
$y'' = -1, tag13$
$y'' + y' - 2y$
$= -1 - x - dfrac12 - 2(-dfrac12x^2 - dfrac12x - dfrac34) = -1 - x - dfrac12 + x^2 + x + dfrac32 = x^2. tag14$
answered Mar 24 at 20:21
Robert LewisRobert Lewis
48.9k23168
$endgroup$
1
$begingroup$
How did you do steps 6, 7, and 8?
$endgroup$
– LuminousNutria
Mar 24 at 20:35
1
$begingroup$
@LuminousNutria: I simply took (5) and equated the coefficients of powers of $x$--$x^0 = 1, ; x^1, ; x^2$--occurring on each side.
$endgroup$
– Robert Lewis
Mar 24 at 20:40
$begingroup$
So, $-2A=1$ because the coefficient of $x^2$ is 1? And $2(A - B) = 0$ because the coefficients of $2x - 2x$ equal $1 - 1$ and are therefore zero? And $(2A+B-2C) = 0$ because it has no powers of x?
$endgroup$
– LuminousNutria
Mar 24 at 20:55
1
$begingroup$
@LuminousNutria: yes! Basically what we're doing is using the fact that each side is a quadratic polynomial in $x$, since they are equal, the coefficients have to be the same. Equal polynomials have equal coefficients! Cheers!
$endgroup$
– Robert Lewis
Mar 24 at 21:05
add a comment |
$begingroup$
With
$y = Ax^2 + Bx + C, tag 1$
we may substitute $y$, $y'$, and $y''$ into
$y'' + y' - 2y = x^2, tag 2$
viz,
$y' = 2Ax + B, tag 3$
$y'' = 2A, tag 4$
$y'' + y' - 2y = 2A + (2Ax + B) - 2(Ax^2 + Bx + C) = x^2; tag 5$
we group together like powers of $x$:
$-2Ax^2 + 2(A - B) x + (2A + B -2C) = x^2, tag 5$
from which we infer
$-2A = 1, tag 6$
$2(A - B) = 0, tag 7$
$2A + B - 2C = 0; tag 8$
thus
$A = B = -dfrac12, tag 9$
$C = -dfrac34, tag10$
and of course
$y = -dfrac12x^2 - dfrac12x - dfrac34. tag11$
We Check:
From (11),
$y' = -x - dfrac12, tag12$
$y'' = -1, tag13$
$y'' + y' - 2y$
$= -1 - x - dfrac12 - 2(-dfrac12x^2 - dfrac12x - dfrac34) = -1 - x - dfrac12 + x^2 + x + dfrac32 = x^2. tag14$
answered Mar 24 at 20:21
Robert LewisRobert Lewis
48.9k23168
$endgroup$
With
$y = Ax^2 + Bx + C, tag 1$
we may substitute $y$, $y'$, and $y''$ into
$y'' + y' - 2y = x^2, tag 2$
viz,
$y' = 2Ax + B, tag 3$
$y'' = 2A, tag 4$
$y'' + y' - 2y = 2A + (2Ax + B) - 2(Ax^2 + Bx + C) = x^2; tag 5$
we group together like powers of $x$:
$-2Ax^2 + 2(A - B) x + (2A + B -2C) = x^2, tag 5$
from which we infer
$-2A = 1, tag 6$
$2(A - B) = 0, tag 7$
$2A + B - 2C = 0; tag 8$
thus
$A = B = -dfrac12, tag 9$
$C = -dfrac34, tag10$
and of course
$y = -dfrac12x^2 - dfrac12x - dfrac34. tag11$
We Check:
From (11),
$y' = -x - dfrac12, tag12$
$y'' = -1, tag13$
$y'' + y' - 2y$
$= -1 - x - dfrac12 - 2(-dfrac12x^2 - dfrac12x - dfrac34) = -1 - x - dfrac12 + x^2 + x + dfrac32 = x^2. tag14$
answered Mar 24 at 20:21
Robert LewisRobert Lewis
48.9k23168
edited Mar 24 at 20:30
answered Mar 24 at 20:21
Robert LewisRobert Lewis
48.9k23168
answered Mar 24 at 20:21
Robert LewisRobert Lewis
48.9k23168
answered Mar 24 at 20:21
Robert LewisRobert Lewis
48.9k23168
48.9k23168
1
$begingroup$
How did you do steps 6, 7, and 8?
$endgroup$
– LuminousNutria
Mar 24 at 20:35
1
$begingroup$
@LuminousNutria: I simply took (5) and equated the coefficients of powers of $x$--$x^0 = 1, ; x^1, ; x^2$--occurring on each side.
$endgroup$
– Robert Lewis
Mar 24 at 20:40
$begingroup$
So, $-2A=1$ because the coefficient of $x^2$ is 1? And $2(A - B) = 0$ because the coefficients of $2x - 2x$ equal $1 - 1$ and are therefore zero? And $(2A+B-2C) = 0$ because it has no powers of x?
$endgroup$
– LuminousNutria
Mar 24 at 20:55
1
$begingroup$
@LuminousNutria: yes! Basically what we're doing is using the fact that each side is a quadratic polynomial in $x$, since they are equal, the coefficients have to be the same. Equal polynomials have equal coefficients! Cheers!
$endgroup$
– Robert Lewis
Mar 24 at 21:05
add a comment |
1
$begingroup$
How did you do steps 6, 7, and 8?
$endgroup$
– LuminousNutria
Mar 24 at 20:35
1
$begingroup$
@LuminousNutria: I simply took (5) and equated the coefficients of powers of $x$--$x^0 = 1, ; x^1, ; x^2$--occurring on each side.
$endgroup$
– Robert Lewis
Mar 24 at 20:40
$begingroup$
So, $-2A=1$ because the coefficient of $x^2$ is 1? And $2(A - B) = 0$ because the coefficients of $2x - 2x$ equal $1 - 1$ and are therefore zero? And $(2A+B-2C) = 0$ because it has no powers of x?
$endgroup$
– LuminousNutria
Mar 24 at 20:55
1
$begingroup$
@LuminousNutria: yes! Basically what we're doing is using the fact that each side is a quadratic polynomial in $x$, since they are equal, the coefficients have to be the same. Equal polynomials have equal coefficients! Cheers!
$endgroup$
– Robert Lewis
Mar 24 at 21:05
1
1
$begingroup$
How did you do steps 6, 7, and 8?
$endgroup$
– LuminousNutria
Mar 24 at 20:35
$begingroup$
How did you do steps 6, 7, and 8?
$endgroup$
– LuminousNutria
Mar 24 at 20:35
1
1
$begingroup$
@LuminousNutria: I simply took (5) and equated the coefficients of powers of $x$--$x^0 = 1, ; x^1, ; x^2$--occurring on each side.
$endgroup$
– Robert Lewis
Mar 24 at 20:40
$begingroup$
@LuminousNutria: I simply took (5) and equated the coefficients of powers of $x$--$x^0 = 1, ; x^1, ; x^2$--occurring on each side.
$endgroup$
– Robert Lewis
Mar 24 at 20:40
$begingroup$
So, $-2A=1$ because the coefficient of $x^2$ is 1? And $2(A - B) = 0$ because the coefficients of $2x - 2x$ equal $1 - 1$ and are therefore zero? And $(2A+B-2C) = 0$ because it has no powers of x?
$endgroup$
– LuminousNutria
Mar 24 at 20:55
$begingroup$
So, $-2A=1$ because the coefficient of $x^2$ is 1? And $2(A - B) = 0$ because the coefficients of $2x - 2x$ equal $1 - 1$ and are therefore zero? And $(2A+B-2C) = 0$ because it has no powers of x?
$endgroup$
– LuminousNutria
Mar 24 at 20:55
1
1
$begingroup$
@LuminousNutria: yes! Basically what we're doing is using the fact that each side is a quadratic polynomial in $x$, since they are equal, the coefficients have to be the same. Equal polynomials have equal coefficients! Cheers!
$endgroup$
– Robert Lewis
Mar 24 at 21:05
$begingroup$
@LuminousNutria: yes! Basically what we're doing is using the fact that each side is a quadratic polynomial in $x$, since they are equal, the coefficients have to be the same. Equal polynomials have equal coefficients! Cheers!
$endgroup$
– Robert Lewis
Mar 24 at 21:05
add a comment |
$begingroup$
Consider $z^2+z-2=(z+2)(z-1)=-2(1+frac12z)(1-z)$. Its inverse (as a formal power series) is
$$
-frac12(1+z+z^2+dotsb)left(1-frac12z+frac14z^2+dotsbright)
=-dfrac12-dfrac14z-dfrac38z^2+dotsb
$$
Interpret $z$ as the differentiation operator and “multiply” by $x^2$ (that's why we can discard higher order terms):
$$
-dfrac12x^2-dfrac142x-dfrac382=-dfrac12x^2-dfrac12x-dfrac34
$$
answered Mar 24 at 20:41
egregegreg
186k1486208
$endgroup$
add a comment |
$begingroup$
Consider $z^2+z-2=(z+2)(z-1)=-2(1+frac12z)(1-z)$. Its inverse (as a formal power series) is
$$
-frac12(1+z+z^2+dotsb)left(1-frac12z+frac14z^2+dotsbright)
=-dfrac12-dfrac14z-dfrac38z^2+dotsb
$$
Interpret $z$ as the differentiation operator and “multiply” by $x^2$ (that's why we can discard higher order terms):
$$
-dfrac12x^2-dfrac142x-dfrac382=-dfrac12x^2-dfrac12x-dfrac34
$$
answered Mar 24 at 20:41
egregegreg
186k1486208
$endgroup$
add a comment |
$begingroup$
Consider $z^2+z-2=(z+2)(z-1)=-2(1+frac12z)(1-z)$. Its inverse (as a formal power series) is
$$
-frac12(1+z+z^2+dotsb)left(1-frac12z+frac14z^2+dotsbright)
=-dfrac12-dfrac14z-dfrac38z^2+dotsb
$$
Interpret $z$ as the differentiation operator and “multiply” by $x^2$ (that's why we can discard higher order terms):
$$
-dfrac12x^2-dfrac142x-dfrac382=-dfrac12x^2-dfrac12x-dfrac34
$$
answered Mar 24 at 20:41
egregegreg
186k1486208
$endgroup$
Consider $z^2+z-2=(z+2)(z-1)=-2(1+frac12z)(1-z)$. Its inverse (as a formal power series) is
$$
-frac12(1+z+z^2+dotsb)left(1-frac12z+frac14z^2+dotsbright)
=-dfrac12-dfrac14z-dfrac38z^2+dotsb
$$
Interpret $z$ as the differentiation operator and “multiply” by $x^2$ (that's why we can discard higher order terms):
$$
-dfrac12x^2-dfrac142x-dfrac382=-dfrac12x^2-dfrac12x-dfrac34
$$
answered Mar 24 at 20:41
egregegreg
186k1486208
answered Mar 24 at 20:41
egregegreg
186k1486208
answered Mar 24 at 20:41
egregegreg
186k1486208
answered Mar 24 at 20:41
egregegreg
186k1486208
186k1486208
add a comment |
add a comment |
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