MAT244--2019F > Term Test 1

Problem 2 (noon)

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Yiheng Bian:
I think in part b when solve y_2 you skip many steps
the equation is
$$
xy_2'-y_2=xcosx-sinx
$$
$$
\int{xy_2'-y_2}=\int{xcos-sinx}
$$
So
$$
xy_2=xsinx
$$
Therefore
$$
y_2=sinx
$$

Nan Yang:
Thanks for your advise. For showing a solution, I need write more details.  BUT I think it is very clear to see that $y_2$ is $sin(x)$  and it can save time for us in writing a test.   ;D

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