Find the higher order derivative of y

\frac{d^2}{dx^2}\left(y\right)+\left(x^2-2x+2\right)\cdot y\cdot\frac{}{}=0

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Answer

$y\left(2-2x+x^2\right)=0$

Step by step solution

Problem

$\frac{d^2}{dx^2}\left(y\right)+\left(x^2-2x+2\right)\cdot y\cdot\frac{}{}=0$
1

Simplifying the fraction by $$

$1y\left(2-2x+x^2\right)+\frac{d^2}{dx^2}\left(y\right)=0$
2

Any expression multiplied by $1$ is equal to itself

$y\left(2-2x+x^2\right)+\frac{d^2}{dx^2}\left(y\right)=0$
3

Rewriting the high order derivative

$y\left(2-2x+x^2\right)+\frac{d^{\left(2-1\right)}}{dx^{\left(2-1\right)}}\left(\frac{d}{dx}\left(y\right)\right)=0$
4

The derivative of the constant function is equal to zero

$y\left(2-2x+x^2\right)+\frac{d^{\left(2-1\right)}}{dx^{\left(2-1\right)}}\left(0\right)=0$
5

Subtract the values $2$ and $-1$

$y\left(2-2x+x^2\right)+\frac{d^{1}}{dx^{1}}\left(0\right)=0$
6

Any expression to the power of $1$ is equal to that same expression

$y\left(2-2x+x^2\right)+\frac{d}{dx}\left(0\right)=0$
7

The derivative of the constant function is equal to zero

$y\left(2-2x+x^2\right)+0=0$
8

$x+0=x$, where $x$ is any expression

$y\left(2-2x+x^2\right)=0$

Answer

$y\left(2-2x+x^2\right)=0$

Problem Analysis

Main topic:

Differential calculus

Time to solve it:

0.28 seconds

Views:

121