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Find values for K
- Thread starter raven720
- Start date
BigGlenntheHeavy
Senior Member
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- Mar 8, 2009
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\(\displaystyle y \ = \ x^{2}+k\)
\(\displaystyle y \ ' \ = \ 2x\)
\(\displaystyle 2(x^{2}+k)-x(2x) \ = \ 10\)
\(\displaystyle 2x^{2}+2k-2x^{2} \ = \ 10\)
\(\displaystyle k \ = \ 5\)
\(\displaystyle y \ ' \ = \ 2x\)
\(\displaystyle 2(x^{2}+k)-x(2x) \ = \ 10\)
\(\displaystyle 2x^{2}+2k-2x^{2} \ = \ 10\)
\(\displaystyle k \ = \ 5\)
Hello, raven720!
\(\displaystyle \text{We have: }\;\begin{array}{c} y \:=\:x^2+k \\ y' \:=\:2x \end{array}\)
\(\displaystyle \text{Substitute into [1]: }\;2\overbrace{(x^2+k)}^{y} - x\overbrace{(2x)}^{y'} \:=\:10 \quad\Rightarrow\quad 2x^2 + 2k - 2x^2 \:=\:10\)
. . . . . . . . . \(\displaystyle 2k \:=\:10 \quad\Rightarrow\quad \boxed{k \:=\:5}\)
\(\displaystyle \text{Find the values of }k\text{ for which }y = x^2 + k\text{ is a solution to: }\: 2y - xy' \:=\: 10\;\;[1]\)
\(\displaystyle \text{We have: }\;\begin{array}{c} y \:=\:x^2+k \\ y' \:=\:2x \end{array}\)
\(\displaystyle \text{Substitute into [1]: }\;2\overbrace{(x^2+k)}^{y} - x\overbrace{(2x)}^{y'} \:=\:10 \quad\Rightarrow\quad 2x^2 + 2k - 2x^2 \:=\:10\)
. . . . . . . . . \(\displaystyle 2k \:=\:10 \quad\Rightarrow\quad \boxed{k \:=\:5}\)