Inhomogeneous differential equation












0














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 $a_0neq 0$.



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  • 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.
    – Kavi Rama Murthy
    Jan 4 at 12:38
















0














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










share|cite|improve this question
























  • 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.
    – Kavi Rama Murthy
    Jan 4 at 12:38














0












0








0







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










share|cite|improve this question















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







differential-equations polynomials






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edited Jan 4 at 12:35









Scientifica

6,37641335




6,37641335










asked Jan 4 at 12:24









Kasper LarsenKasper Larsen

113




113












  • 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.
    – Kavi Rama Murthy
    Jan 4 at 12:38


















  • 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.
    – Kavi Rama Murthy
    Jan 4 at 12:38
















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.
– Kavi Rama Murthy
Jan 4 at 12:38




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.
– Kavi Rama Murthy
Jan 4 at 12:38










1 Answer
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Just plug in $q(x)=ax+b$ in the equation and determine $a$ and $b$ by comparing coefficients.. The answer is $q(x)=frac {alpha} {a_0} x+frac {beta a_0 -alpha a_1} {a_0^{2}}$.






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  • I understand it now, thank you!
    – Kasper Larsen
    Jan 4 at 13:08













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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









0














Just plug in $q(x)=ax+b$ in the equation and determine $a$ and $b$ by comparing coefficients.. The answer is $q(x)=frac {alpha} {a_0} x+frac {beta a_0 -alpha a_1} {a_0^{2}}$.






share|cite|improve this answer





















  • I understand it now, thank you!
    – Kasper Larsen
    Jan 4 at 13:08


















0














Just plug in $q(x)=ax+b$ in the equation and determine $a$ and $b$ by comparing coefficients.. The answer is $q(x)=frac {alpha} {a_0} x+frac {beta a_0 -alpha a_1} {a_0^{2}}$.






share|cite|improve this answer





















  • I understand it now, thank you!
    – Kasper Larsen
    Jan 4 at 13:08
















0












0








0






Just plug in $q(x)=ax+b$ in the equation and determine $a$ and $b$ by comparing coefficients.. The answer is $q(x)=frac {alpha} {a_0} x+frac {beta a_0 -alpha a_1} {a_0^{2}}$.






share|cite|improve this answer












Just plug in $q(x)=ax+b$ in the equation and determine $a$ and $b$ by comparing coefficients.. The answer is $q(x)=frac {alpha} {a_0} x+frac {beta a_0 -alpha a_1} {a_0^{2}}$.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Jan 4 at 12:33









Kavi Rama MurthyKavi Rama Murthy

51.8k32055




51.8k32055












  • I understand it now, thank you!
    – Kasper Larsen
    Jan 4 at 13:08




















  • I understand it now, thank you!
    – Kasper Larsen
    Jan 4 at 13:08


















I understand it now, thank you!
– Kasper Larsen
Jan 4 at 13:08






I understand it now, thank you!
– Kasper Larsen
Jan 4 at 13:08




















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