Differential Equation.

[sepoto]

My question is about b). I follow b) most of the way through to the point where:

\(\displaystyle y+1=(3/2+\sqrt{2})^2\)

My question starts where the solutions manual says:
\(\displaystyle y+1=(3/2+\sqrt{2})^2=13/2+3\sqrt{2}\)

For some reason my algebra is coming out different. I have:
\(\displaystyle y=(3/2+\sqrt{2})^2-1=9/4+8/4-4/4=-13/4\)

Thanks in advance for any help on this problem...

 

[mathmari]

\(\displaystyle y=(\frac{3}{2}+\sqrt{2})^2-1=(\frac{3}{2})^2+2 \cdot \frac{3}{2} \sqrt{2} + (\sqrt{2})^2-1=\frac{9}{4}+3 \sqrt{2}+2-1=\frac{9}{4}+3 \sqrt{2}+1=\frac{9}{4}+3 \sqrt{2} +\frac{4}{4}=\frac{13}{4}+3 \sqrt{2}\)