Inhomogeneous differential equation
$begingroup$
How would you solve this:
Let $p(x)=alpha x+beta$ be a first degree polynomial where $alpha$ and $beta$ are arbitrary real numbers. Show that there is a first degree polynomial $q$ that solves the inhomogeneous differential equation $a_ny^{(n)}+a_{n-1}y^{(n-1)}+a_{n-2}y^{(n-2)}+...+a_1y'+a_0y=p$, where $a_0neq 0$.
Thanks on beforehand
ordinary-differential-equations polynomials
edited Jan 4 at 12:35
Scientifica
6,75141335
asked Jan 4 at 12:24
Kasper LarsenKasper Larsen
113
$endgroup$
add a comment |
$begingroup$
How would you solve this:
Let $p(x)=alpha x+beta$ be a first degree polynomial where $alpha$ and $beta$ are arbitrary real numbers. Show that there is a first degree polynomial $q$ that solves the inhomogeneous differential equation $a_ny^{(n)}+a_{n-1}y^{(n-1)}+a_{n-2}y^{(n-2)}+...+a_1y'+a_0y=p$, where $a_0neq 0$.
Thanks on beforehand
ordinary-differential-equations polynomials
edited Jan 4 at 12:35
Scientifica
6,75141335
asked Jan 4 at 12:24
Kasper LarsenKasper Larsen
113
$endgroup$
$begingroup$
For a first degree polynomial $y$ the derivatives $y^{(k)}$ are all $0$ for $k >1$. So you are really working with a first order DE.
$endgroup$
– Kavi Rama Murthy
Jan 4 at 12:38
add a comment |
$begingroup$
How would you solve this:
Let $p(x)=alpha x+beta$ be a first degree polynomial where $alpha$ and $beta$ are arbitrary real numbers. Show that there is a first degree polynomial $q$ that solves the inhomogeneous differential equation $a_ny^{(n)}+a_{n-1}y^{(n-1)}+a_{n-2}y^{(n-2)}+...+a_1y'+a_0y=p$, where $a_0neq 0$.
Thanks on beforehand
ordinary-differential-equations polynomials
edited Jan 4 at 12:35
Scientifica
6,75141335
asked Jan 4 at 12:24
Kasper LarsenKasper Larsen
113
$endgroup$
How would you solve this:
Let $p(x)=alpha x+beta$ be a first degree polynomial where $alpha$ and $beta$ are arbitrary real numbers. Show that there is a first degree polynomial $q$ that solves the inhomogeneous differential equation $a_ny^{(n)}+a_{n-1}y^{(n-1)}+a_{n-2}y^{(n-2)}+...+a_1y'+a_0y=p$, where $a_0neq 0$.
Thanks on beforehand
ordinary-differential-equations polynomials
ordinary-differential-equations polynomials
edited Jan 4 at 12:35
Scientifica
6,75141335
asked Jan 4 at 12:24
Kasper LarsenKasper Larsen
113
edited Jan 4 at 12:35
Scientifica
6,75141335
asked Jan 4 at 12:24
Kasper LarsenKasper Larsen
113
edited Jan 4 at 12:35
Scientifica
6,75141335
edited Jan 4 at 12:35
Scientifica
6,75141335
edited Jan 4 at 12:35
Scientifica
6,75141335
6,75141335
asked Jan 4 at 12:24
Kasper LarsenKasper Larsen
113
asked Jan 4 at 12:24
Kasper LarsenKasper Larsen
113
asked Jan 4 at 12:24
Kasper LarsenKasper Larsen
113
113
$begingroup$
For a first degree polynomial $y$ the derivatives $y^{(k)}$ are all $0$ for $k >1$. So you are really working with a first order DE.
$endgroup$
– Kavi Rama Murthy
Jan 4 at 12:38
add a comment |
$begingroup$
For a first degree polynomial $y$ the derivatives $y^{(k)}$ are all $0$ for $k >1$. So you are really working with a first order DE.
$endgroup$
– Kavi Rama Murthy
Jan 4 at 12:38
$begingroup$
For a first degree polynomial $y$ the derivatives $y^{(k)}$ are all $0$ for $k >1$. So you are really working with a first order DE.
$endgroup$
– Kavi Rama Murthy
Jan 4 at 12:38
$begingroup$
For a first degree polynomial $y$ the derivatives $y^{(k)}$ are all $0$ for $k >1$. So you are really working with a first order DE.
$endgroup$
– Kavi Rama Murthy
Jan 4 at 12:38
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Just plug in $q(x)=ax+b$ in the equation and determine $a$ and $b$ by comparing coefficients.. The answer is $q(x)=frac {alpha} {a_0} x+frac {beta a_0 -alpha a_1} {a_0^{2}}$.
answered Jan 4 at 12:33
Kavi Rama MurthyKavi Rama Murthy
57.6k42160
$endgroup$
$begingroup$
I understand it now, thank you!
$endgroup$
– Kasper Larsen
Jan 4 at 13:08
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Just plug in $q(x)=ax+b$ in the equation and determine $a$ and $b$ by comparing coefficients.. The answer is $q(x)=frac {alpha} {a_0} x+frac {beta a_0 -alpha a_1} {a_0^{2}}$.
answered Jan 4 at 12:33
Kavi Rama MurthyKavi Rama Murthy
57.6k42160
$endgroup$
$begingroup$
I understand it now, thank you!
$endgroup$
– Kasper Larsen
Jan 4 at 13:08
add a comment |
$begingroup$
Just plug in $q(x)=ax+b$ in the equation and determine $a$ and $b$ by comparing coefficients.. The answer is $q(x)=frac {alpha} {a_0} x+frac {beta a_0 -alpha a_1} {a_0^{2}}$.
answered Jan 4 at 12:33
Kavi Rama MurthyKavi Rama Murthy
57.6k42160
$endgroup$
$begingroup$
I understand it now, thank you!
$endgroup$
– Kasper Larsen
Jan 4 at 13:08
add a comment |
$begingroup$
Just plug in $q(x)=ax+b$ in the equation and determine $a$ and $b$ by comparing coefficients.. The answer is $q(x)=frac {alpha} {a_0} x+frac {beta a_0 -alpha a_1} {a_0^{2}}$.
answered Jan 4 at 12:33
Kavi Rama MurthyKavi Rama Murthy
57.6k42160
$endgroup$
Just plug in $q(x)=ax+b$ in the equation and determine $a$ and $b$ by comparing coefficients.. The answer is $q(x)=frac {alpha} {a_0} x+frac {beta a_0 -alpha a_1} {a_0^{2}}$.
answered Jan 4 at 12:33
Kavi Rama MurthyKavi Rama Murthy
57.6k42160
answered Jan 4 at 12:33
Kavi Rama MurthyKavi Rama Murthy
57.6k42160
answered Jan 4 at 12:33
Kavi Rama MurthyKavi Rama Murthy
57.6k42160
answered Jan 4 at 12:33
Kavi Rama MurthyKavi Rama Murthy
57.6k42160
57.6k42160
$begingroup$
I understand it now, thank you!
$endgroup$
– Kasper Larsen
Jan 4 at 13:08
add a comment |
$begingroup$
I understand it now, thank you!
$endgroup$
– Kasper Larsen
Jan 4 at 13:08
$begingroup$
I understand it now, thank you!
$endgroup$
– Kasper Larsen
Jan 4 at 13:08
$begingroup$
I understand it now, thank you!
$endgroup$
– Kasper Larsen
Jan 4 at 13:08
add a comment |
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$begingroup$
For a first degree polynomial $y$ the derivatives $y^{(k)}$ are all $0$ for $k >1$. So you are really working with a first order DE.
$endgroup$
– Kavi Rama Murthy
Jan 4 at 12:38