Find the equation of the curve $y''=2x+1$
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I am stuck on the last step of the following equation:
If $y''=2x+1$ and there is a stationary point at $(3,2)$ find the equation of the curve.
So far I have the following:
If $y''=2x+1$
$y'=x^2+x+c$
then use the given point to solve for $c$ as follows:
$0=3^2+3+c$
$c=-12$
so, if $y'=x^2+x-12$
$y= frac{x^3}{3}+frac{x^2}{2}-12x+c$
now I know to find the final equation of the curve I need to find the last $c$ but am stumped on how to do so? any help would be appreciated!
differential-equations curves
edited Nov 18 at 10:56
Kelvin Lois
3,0872823
asked Nov 18 at 8:32
Holly Millican
528
add a comment |
up vote
0
down vote
favorite
I am stuck on the last step of the following equation:
If $y''=2x+1$ and there is a stationary point at $(3,2)$ find the equation of the curve.
So far I have the following:
If $y''=2x+1$
$y'=x^2+x+c$
then use the given point to solve for $c$ as follows:
$0=3^2+3+c$
$c=-12$
so, if $y'=x^2+x-12$
$y= frac{x^3}{3}+frac{x^2}{2}-12x+c$
now I know to find the final equation of the curve I need to find the last $c$ but am stumped on how to do so? any help would be appreciated!
differential-equations curves
edited Nov 18 at 10:56
Kelvin Lois
3,0872823
asked Nov 18 at 8:32
Holly Millican
528
It should pass through the given point
– Semsem
Nov 18 at 8:43
but the given point is a stationary point, so wouldn't it only apply to the first derivative?
– Holly Millican
Nov 18 at 8:45
1
Use $y(3)=2$ for determining the constant
– Fakemistake
Nov 18 at 8:46
thank you, that got me where I needed
– Holly Millican
Nov 18 at 8:52
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I am stuck on the last step of the following equation:
If $y''=2x+1$ and there is a stationary point at $(3,2)$ find the equation of the curve.
So far I have the following:
If $y''=2x+1$
$y'=x^2+x+c$
then use the given point to solve for $c$ as follows:
$0=3^2+3+c$
$c=-12$
so, if $y'=x^2+x-12$
$y= frac{x^3}{3}+frac{x^2}{2}-12x+c$
now I know to find the final equation of the curve I need to find the last $c$ but am stumped on how to do so? any help would be appreciated!
differential-equations curves
edited Nov 18 at 10:56
Kelvin Lois
3,0872823
asked Nov 18 at 8:32
Holly Millican
528
I am stuck on the last step of the following equation:
If $y''=2x+1$ and there is a stationary point at $(3,2)$ find the equation of the curve.
So far I have the following:
If $y''=2x+1$
$y'=x^2+x+c$
then use the given point to solve for $c$ as follows:
$0=3^2+3+c$
$c=-12$
so, if $y'=x^2+x-12$
$y= frac{x^3}{3}+frac{x^2}{2}-12x+c$
now I know to find the final equation of the curve I need to find the last $c$ but am stumped on how to do so? any help would be appreciated!
differential-equations curves
differential-equations curves
edited Nov 18 at 10:56
Kelvin Lois
3,0872823
asked Nov 18 at 8:32
Holly Millican
528
edited Nov 18 at 10:56
Kelvin Lois
3,0872823
asked Nov 18 at 8:32
Holly Millican
528
edited Nov 18 at 10:56
Kelvin Lois
3,0872823
edited Nov 18 at 10:56
Kelvin Lois
3,0872823
edited Nov 18 at 10:56
Kelvin Lois
3,0872823
3,0872823
asked Nov 18 at 8:32
Holly Millican
528
asked Nov 18 at 8:32
Holly Millican
528
asked Nov 18 at 8:32
Holly Millican
528
528
It should pass through the given point
– Semsem
Nov 18 at 8:43
but the given point is a stationary point, so wouldn't it only apply to the first derivative?
– Holly Millican
Nov 18 at 8:45
1
Use $y(3)=2$ for determining the constant
– Fakemistake
Nov 18 at 8:46
thank you, that got me where I needed
– Holly Millican
Nov 18 at 8:52
add a comment |
It should pass through the given point
– Semsem
Nov 18 at 8:43
but the given point is a stationary point, so wouldn't it only apply to the first derivative?
– Holly Millican
Nov 18 at 8:45
1
Use $y(3)=2$ for determining the constant
– Fakemistake
Nov 18 at 8:46
thank you, that got me where I needed
– Holly Millican
Nov 18 at 8:52
It should pass through the given point
– Semsem
Nov 18 at 8:43
It should pass through the given point
– Semsem
Nov 18 at 8:43
but the given point is a stationary point, so wouldn't it only apply to the first derivative?
– Holly Millican
Nov 18 at 8:45
but the given point is a stationary point, so wouldn't it only apply to the first derivative?
– Holly Millican
Nov 18 at 8:45
1
1
Use $y(3)=2$ for determining the constant
– Fakemistake
Nov 18 at 8:46
Use $y(3)=2$ for determining the constant
– Fakemistake
Nov 18 at 8:46
thank you, that got me where I needed
– Holly Millican
Nov 18 at 8:52
thank you, that got me where I needed
– Holly Millican
Nov 18 at 8:52
add a comment |
1 Answer
1
active
oldest
votes
up vote
1
down vote
It would be better to tackle the problem this way: Integrating $y''(x)=2x+1$ twice to obtain begin{align}y'(x)&=x^2+x+c\y(x)&=frac{1}{3}x^3+frac{1}{2}x^2+cx+dend{align}
From $y'(3)=0$ you get $c=-12$ and from $y(3)=2$ you get $frac{49}{2}$.
answered Nov 18 at 9:06
Fakemistake
1,660815
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
It would be better to tackle the problem this way: Integrating $y''(x)=2x+1$ twice to obtain begin{align}y'(x)&=x^2+x+c\y(x)&=frac{1}{3}x^3+frac{1}{2}x^2+cx+dend{align}
From $y'(3)=0$ you get $c=-12$ and from $y(3)=2$ you get $frac{49}{2}$.
answered Nov 18 at 9:06
Fakemistake
1,660815
add a comment |
up vote
1
down vote
It would be better to tackle the problem this way: Integrating $y''(x)=2x+1$ twice to obtain begin{align}y'(x)&=x^2+x+c\y(x)&=frac{1}{3}x^3+frac{1}{2}x^2+cx+dend{align}
From $y'(3)=0$ you get $c=-12$ and from $y(3)=2$ you get $frac{49}{2}$.
answered Nov 18 at 9:06
Fakemistake
1,660815
add a comment |
up vote
1
down vote
up vote
1
down vote
It would be better to tackle the problem this way: Integrating $y''(x)=2x+1$ twice to obtain begin{align}y'(x)&=x^2+x+c\y(x)&=frac{1}{3}x^3+frac{1}{2}x^2+cx+dend{align}
From $y'(3)=0$ you get $c=-12$ and from $y(3)=2$ you get $frac{49}{2}$.
answered Nov 18 at 9:06
Fakemistake
1,660815
It would be better to tackle the problem this way: Integrating $y''(x)=2x+1$ twice to obtain begin{align}y'(x)&=x^2+x+c\y(x)&=frac{1}{3}x^3+frac{1}{2}x^2+cx+dend{align}
From $y'(3)=0$ you get $c=-12$ and from $y(3)=2$ you get $frac{49}{2}$.
answered Nov 18 at 9:06
Fakemistake
1,660815
answered Nov 18 at 9:06
Fakemistake
1,660815
answered Nov 18 at 9:06
Fakemistake
1,660815
answered Nov 18 at 9:06
Fakemistake
1,660815
1,660815
add a comment |
add a comment |
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It should pass through the given point
– Semsem
Nov 18 at 8:43
but the given point is a stationary point, so wouldn't it only apply to the first derivative?
– Holly Millican
Nov 18 at 8:45
1
Use $y(3)=2$ for determining the constant
– Fakemistake
Nov 18 at 8:46
thank you, that got me where I needed
– Holly Millican
Nov 18 at 8:52