Question

(1 point) Solve the initial value problem 10 10(+ 1) My – by = 241, 24t. for t> -1 with y(0) = 3. y=

Answer 1

Given 10(6+1) -6y= 24,Y(0) = 3–(1) On di vi ding with '10(t+1)', weget, dy 6 ,- 246 dt 10(t+1)*10(t+1) dy_ 3__y= 124, which is alinear differential Pdt 5(t+1)5(t+1) equation of first orderin y' and it is of the form + P(x)y=06). here Pon=-518+1) • 20) = 31:1) Integrating factor, I.F.,u(t) = eStw* es " = 3hca} = (+1)*(* nina = In(a") The general solution of (1) is y.l.F = ſect), 1.F di +c, where c is an arbitrary constant. = y(+1)* = |(5141; }(+13 di +6 = *(1+1)* =*/ 07(6+1)dt+e (* 1 *dx = -+-1) = x(+1)*-* ( 0, (+3)6+03-01-(3)le+13)+c = (+1)ỉ ?( 3 ) (+) *-(( 8 ):+1) + = *(1+1)} = (-41)+13+(*)te-+18)+c =»(+1)} = (-4f6+1)+10(1+1)+c =y=(-44 +10(8 +1)+c1+113 →(2),is the general solution of (1). Applying the initial condition y(0) = 3, we get 3=(-0 +10)+c =c=-7 :: y=(-48 +10(t+1))– 7(6+1)$, is the solution of (1).