Question

find the value of k for which \[y=x^{2}+k\] is a solution to the differential equation:
\[2y-xy'=28\]
my work below

Answer 1

\[2y-x'y=28\rightarrow x\frac{dy}{dx}=2y-28\rightarrow \]
\[\frac{1}{2}\int\limits_{}^{}\frac{ dy }{ y-14 }=\int\limits\frac{dx}{x}\rightarrow \]
\[y=\frac{x+c+14}{e^{\frac{1}{2}}}\]
this is where I get confused.

Answer 2

err it will be easier if u go the OTHER way....by differeniating y....

Answer 3

uh ok...

Answer 4

y'=2x n put it back in the diff eqn...

Answer 5

oh it should be
y'=2x
AND
y=x^2 + k
into the DE

Answer 6

so the answer is 14

Answer 7

@Kkutie7 correct - good job! :)