How to show that y=C1e^x+C2x is a solution for y"(1-x)+y'x-y=0?

Question

john_es · Answer

I think the direct way: calculate y'' and y', then plug in into the differential equation. When you sum all terms the result should be zero.

imtant · Answer

Yes,but do you know how to differentiate the y?

john_es · Answer

For the first derivative (C1 and C2 are constants),
$$y'=(C_1e^x+C_2x)'=C_1e^x+C_2$$I think you can follow from this point.

john_es · Answer

Do you understand how to finish the problem?

imtant · Answer

No.Can you show how you factor the x?

john_es · Answer

What do you mean by "factor the x"?

imtant · Answer

I don't understand how you get C1e^x+C2.

john_es · Answer

Ah, ok. It's all in the derivatives,
$$y'=(C_1e^2+C_2x)'=(C_1e^x)'+(C_2x)'=C_1e^x+C_2$$
Remembering the derivative of x is 1.

john_es · Answer

The second derivative,
$$y''=C_1e^x$$
And then you can plug in into the original differential equation,
$$(C_1e^x)(1-x)+(C_1e^x+C_2)x-(C_1e^x+C_2x)$$

imtant · Answer

Okay.So for the second differentiate there's no need to write C2? right?

john_es · Answer

right, because (C2)'=0.

imtant · Answer

Thanks John_ES. :)

john_es · Answer

You're welcome.