Question

1 point) the solution of a certain differential equation is of the form y(t)=aexp(3t)+bexp(4t), y ( t ) = a exp ⁡ ( 3 t ) + b exp ⁡ ( 4 t ) , where a a and b b are constants. the solution has initial conditions y(0)=3 y ( 0 ) = 3 and y′(0)=3. y ′ ( 0 ) = 3 . find the solution by using the initial conditions to get linear equations for a a and

Answer 1

y(t) = a exp(3t) + b exp(4t) conditions, y(0) = 3 y'(0) = 3 y(0) = a exp(3 x 0) + b exp(4 x 0) = a exp(0) + b exp(0) = (a x 1) + (b x 1) = a + b y'(0) = 0 so the linear equation is, a + b = 3