5. Solve the given initial value problem 4y" – 4y' – 3y = 0, y(0) = 1, y(0) = 5

Answer:y = [tex]\frac{-7}{4}e^{\frac{-1}{2}x}+ \frac{11}{4}e^{\frac{3}{2}x}[/tex]Step-by-step explanation:Given:4y" – 4y' – 3y = 0, y(0) = 1, y'(0) = 5Now,The auxiliary equation will be given as:4m² - 4m - 3 = 0or4m² + 2m - 6m - 3 = 0or2m (2 + 1) - 3 (2m + 1) = 0or( 2m - 3 ) × ( 2m + 1 ) = 0therefore, m = [tex]\frac{\textup{3}}{\textup{2}}[/tex] and m = [tex]\frac{\textup{-1}}{\textup{2}}[/tex] thus,the general equation comes as:y = [tex]C_1e^{mx}+ C_2e^{mx}[/tex]ory = [tex]C_1e^{\frac{-1}{2}x}+ C_2e^{\frac{3}{2}x}[/tex]now,at y(0) = 1therefore,1 = [tex]C_1e^{\frac{-1}{2}\times0}+ C_2e^{\frac{3}{2}\times0}[/tex]orC₁ + C₂ = 1  .............(1)and, y' = [tex]\frac{-1}{2}C_1e^{\frac{-1}{2}x}+ \frac{3}{2}C_2e^{\frac{3}{2}x}[/tex]also,y'(0) = 5thus,5 = [tex]\frac{-1}{2}C_1e^{\frac{-1}{2}\times0}+ \frac{3}{2}C_2e^{\frac{3}{2}\times0}[/tex]or3C₂ - C₁ = 10 ...........(2)on adding 1 and 2, we get      C₁ + C₂ = 1 + (- C₁ + 3C₂) = 10===============4C₂ = 11 orC₂ = [tex]\frac{\textup{11}}{\textup{4}}[/tex] thus,    C₁ + C₂ = 1 or    C₁ + [tex]\frac{\textup{11}}{\textup{4}}[/tex]  = 1 orC₁ =  [tex]\frac{\textup{-7}}{\textup{4}}[/tex] Hence,The solution is y = [tex]C_1e^{mx}+ C_2e^{mx}[/tex] on substituting the values, y = [tex]\frac{-7}{4}e^{\frac{-1}{2}x}+ \frac{11}{4}e^{\frac{3}{2}x}[/tex]