Question

solve 4y"-25y' = 0

• A. $$C_1-C_2e^{\frac{25}{4}x}$$
• B. $$C_1+C_2e^{\frac{25}{4}x}$$
• C. $$C_2e^{\frac{25}{4}x}$$
• D. $$C_1+C_2e^{\frac{4}{25}x}$$

Answer 1

Answer:

$$C_1+C_2e^{\frac{25}{4}x}$$

4y"-25y' = 0

A.E => 4m² - 25m = 0

m(4m-25m)=0

m=0 ; 4m = 25

m=25/4

$y= C_1 e^{m_1x}+C_2e^{m_2x}$

$=C_1e^{0x}+C_2e^{\frac{25}{4}x}$

$=C_1+C_2e^{\frac{25}{4}x}$