$x{f}^{\u2033}(x)+f(x)={x}^{2}$

Karsyn Beltran
2022-08-01
Answered

Find a polynomial f(x) that satisfies this equation:

$x{f}^{\u2033}(x)+f(x)={x}^{2}$

$x{f}^{\u2033}(x)+f(x)={x}^{2}$

You can still ask an expert for help

esbalatzaj

Answered 2022-08-02
Author has **15** answers

Look for a solution of the form:

$f(x)=\sum _{r=0}^{n}{a}_{r}{x}^{r}$

For some set of constants ${a}_{r}$

Then ${f}^{\prime}(x)=\sum _{r=1}^{n}{a}_{r}r{x}^{r-1}$

${f}^{\u2033}(x)=\sum _{r=2}^{n}{a}_{r}r(r-1){x}^{r-2}$

So we have

$x\sum _{r=2}^{n}{a}_{r}r(r-1){x}^{r-2}+\sum _{r=0}^{n}{a}_{r}{x}^{r}={x}^{2}$

Equating like terms on lhs and rhs we see that:

$6{a}_{3}+{a}_{2}=1$ (for the ${x}^{2}$ terms)

$2{a}_{2}+{a}_{1}=0$ (for the x terms)

${a}_{0}=0$ (for the constant terms)

But we must have ${a}_{2}=1$ for the squared termsbecause there's just ${x}^{2}$ on rhs

Hence ${a}_{1}=-2$

${a}_{3}=0$

So $f(x)={x}^{2}-2x$

$f(x)=\sum _{r=0}^{n}{a}_{r}{x}^{r}$

For some set of constants ${a}_{r}$

Then ${f}^{\prime}(x)=\sum _{r=1}^{n}{a}_{r}r{x}^{r-1}$

${f}^{\u2033}(x)=\sum _{r=2}^{n}{a}_{r}r(r-1){x}^{r-2}$

So we have

$x\sum _{r=2}^{n}{a}_{r}r(r-1){x}^{r-2}+\sum _{r=0}^{n}{a}_{r}{x}^{r}={x}^{2}$

Equating like terms on lhs and rhs we see that:

$6{a}_{3}+{a}_{2}=1$ (for the ${x}^{2}$ terms)

$2{a}_{2}+{a}_{1}=0$ (for the x terms)

${a}_{0}=0$ (for the constant terms)

But we must have ${a}_{2}=1$ for the squared termsbecause there's just ${x}^{2}$ on rhs

Hence ${a}_{1}=-2$

${a}_{3}=0$

So $f(x)={x}^{2}-2x$

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