How to explain without calculation that yp(x) = ax is not a particular solution of 2(x^2)y'' + xy' - y = 10

[Phaedrus]

I came upon this question and have no idea how to explain it, can anyone help?

 

[Steven G]

Try doing it with calculations and see what you get.
Hint: y=ax, y'=a and y" = 0.

 

[Phaedrus]

Try doing it with calculations and see what you get.
Hint: y=ax, y'=a and y" = 0.

But my lecturer wants us to explain without calculations. Is it that I can see how to explain by looking at the calculations?

 

[Steven G]

You are not seeing it by inspection, correct? So try solving it by doing the computation and then see if you can then understand that you did not need to do the computation.

What types of results do you get when you subtract two terms that are multiples of x's (like 9x-2x)?

 

[Phaedrus]

You are not seeing it by inspection, correct? So try solving it by doing the computation and then see if you can then understand that you did not need to do the computation.

What types of results do you get when you subtract two terms that are multiples of x's (like 9x-2x)?

i did the calculations and got the particular solution to be -20. Is the degree of x in the particular solution be the same as the r(x)? If theres no x in the r(x) before preliminary division, the particular solution will not have any x too?

 

[Phaedrus]

Try doing it with calculations and see what you get.
Hint: y=ax, y'=a and y" = 0.

I followed the hints and calculated 0 = 10, which proves that the particular solution ax does not satisfy the given DE, so how can i explain based on this?

 

[Steven G]

If y=ax, you should easily see that y'=a and y"=0.
So 2x^2y" = 0, xy'=ax and y=ax. Then 2x^2y" + xy' - y = 0 + ax -ax =0. This can be seen without writing anything down.

 

[nasi112]

I followed the hints and calculated 0 = 10, which proves that the particular solution ax does not satisfy the given DE, so how can i explain based on this?

If the differential equation is equal to a constant, the particular solution must also be a constant while \(\displaystyle y_p= ax\) is not a constant.

 

[Steven G]

If the differential equation is equal to a constant, the particular solution must also be a constant while \(\displaystyle y_p= ax\) is not a constant.

But y=ax does yield a constant in that differential equation.

 

[LCKurtz]

It seems like a strange question to me. One answer might be that if y = ax then every term on the left side would have a factor of x while the right side doesn't. Whether one would know that without doing "any calculations" is debatable.