Question

Verify by differentiation and substitution that the given function is a solution.
y'= (1/t)y+t y=3t+t^2

Answer 1

is the ' on the y on accident?

Answer 2

\[\Large\rm \color{orangered}{y'}=\frac{1}{t}\color{royalblue}{y}+t\]

So we have,\[\Large\rm \color{royalblue}{y=3t+t^2}\]Differentiating this function gives,\[\Large\rm \color{orangered}{y'=3+2t}\]yes?

So plug these two pieces into the original Differential Equation and verify! :)

Answer 3

@zepdrix so it will be y'= 1/t(3+2t)+t

Answer 4

Woops, you plugged your \(\Large\rm \color{orangered}{y'}\) in for your y by mistake I think:\[\Large\rm \color{orangered}{y'}=\frac{1}{t}\color{royalblue}{(3t+t^2)}+t\]We want to plug \(\Large\rm \color{royalblue}{y}\) into the right side.

Answer 5

I thought the colors would help -_- Hmm..\[\Large\rm \color{orangered}{y'}=\frac{1}{t}\color{royalblue}{y}+t\]\[\Large\rm \color{orangered}{3+2t}=\frac{1}{t}\color{royalblue}{(3t+t^2)}+t\]

Answer 6

so now do I take the derivative or simplify?

Answer 7

Simplify,
show that the left side is equal to the right side.

Answer 8

I simplify and the left side is equal to the right side. Is that all I have to do?

Answer 9

Yes, that verifies that the y we used IS in fact a solution to the differential equation.

Answer 10

It solves it, makes it true.

Answer 11

thank you!!