Find values for K

[raven720]

Find the values for k for which y = x^2 + k is a solution to 2y - xy' = 10

I dont think that it is to hard of a question, but for what ever reason i cant remember how to do this, any help?

 

[BigGlenntheHeavy]

$\displaystyle y \ = \ x^{2}+k$

$\displaystyle y \ ' \ = \ 2x$

$\displaystyle 2(x^{2}+k)-x(2x) \ = \ 10$

$\displaystyle 2x^{2}+2k-2x^{2} \ = \ 10$

$\displaystyle k \ = \ 5$

 

[soroban]

Hello, raven720!

$\displaystyle \text{Find the values of }k\text{ for which }y = x^2 + k\text{ is a solution to: }\: 2y - xy' \:=\: 10\;\;[1]$

$\displaystyle \text{We have: }\;\begin{array}{c} y \:=\:x^2+k \\ y' \:=\:2x \end{array}$

$\displaystyle \text{Substitute into [1]: }\;2\overbrace{(x^2+k)}^{y} - x\overbrace{(2x)}^{y'} \:=\:10 \quad\Rightarrow\quad 2x^2 + 2k - 2x^2 \:=\:10$

. . . . . . . . . $\displaystyle 2k \:=\:10 \quad\Rightarrow\quad \boxed{k \:=\:5}$

 

[raven720]

Thanks for the help, i knew it easier then i thought lol