Assuming ideal op-amplifiers, I did the following steps:
\$v_{out1}=v_1\frac{R_2}{R_1}=2v_1\$
\$v_{out2}=v_o=\left(\frac{v_{out1}}{R_3}+\frac{v_2}{R_4}\right)R_5=4v_1+2v_2\$
My textbook claims \$4v_1-2v_2\$
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Sign up to join this communityAssuming ideal op-amplifiers, I did the following steps:
\$v_{out1}=v_1\frac{R_2}{R_1}=2v_1\$
\$v_{out2}=v_o=\left(\frac{v_{out1}}{R_3}+\frac{v_2}{R_4}\right)R_5=4v_1+2v_2\$
My textbook claims \$4v_1-2v_2\$
These stages are inverting. Your equations are both missing a - in the transfer function.
V1 gets inverted twice, that's why it has a positive coefficient. V2 gets inverted once, that's why it has a negative coefficient.