Writing Equation in Normal Form- Where did the t go?

Question

I've been asked to write the equation $x''+x'+x=0$ in normal form.

Subbing in $x(t) = e^{v(t)}y(t)$ into the diff eq, I obtain $y''+(2v'+1)y'+(v''+(v')^2+v'+1)y=0$.

Eliminating the first order term, we have $2v'+1=0 \implies v' = -\frac{1}{2} \implies v= -\frac{1}{2}t$.

Then $y''+(-\frac{1}{2}t)^2-\frac{1}{2}+1)y=0^{**}$, but my book says that the sol'n is $y''+\frac{3}{4}y=0$. Where did the $t$ in $^{**}$ go?

Answer 1

If $v(t)=-t/2$, there is no $t$ is any derivative of $v$. As $v=-1/2$, you get $$ y''+(0) y'+(0+(-1/2)^2-1/2+1)y =y''+\frac34 y=0 $$

Answer 2

You were almost there $$y''+(2v'+1)y'+(v''+(v')^2+v'+1)y=0$$ Since $2v'+1=0$ $$y''+(v''+(v')^2+v'+1)y=0$$ You have also that $ v' = -\frac{1}{2} \implies v''=0$ $$y''+(0+ \frac 14 -\frac 12+1)y=0$$ Finally: $$y''+\frac34y=0$$