MAT244--2019F > Term Test 1

Problem 2 (noon)

**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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