Answer:
[tex]y(1) = 0.5518[/tex]
Step-by-step explanation:
Given the differential equation [tex]e^{t} \frac{dy}{dt} +e^{t}y = t[/tex]
From the equation;
[tex]e^{t} (\frac{dy}{dt} +y) = t\\\frac{dy}{dt} +y = \frac{t}{e^{t}} \\[/tex]
The resulting equation is a first order differential equation in the form
dy/dt + p(t)y = q(t)
The solution to the DE will be in the form yI = [tex]\int\limits q(t)*I\, dt[/tex] where I is the integrating factor expressed as [tex]I = e^{\int\limits p(t) \, dt }[/tex]
From the DE above p(t ) = 1 and q(t) = [tex]\frac{t}{e^{t} }[/tex]
[tex]I = e^{\int\limits1 \, dt }\\I = e^{t }\\[/tex]
The solution to the DE will become
[tex]y e^{t } = \int\limits e^{t } *\frac{t}{e^{t } } \, dt\\y e^{t } = \int\limits{t} \, dt \\y e^{t } = \frac{t^{2} }{2} + C[/tex]
If y(0) = 1 then;
[tex]1 e^{0 } = \frac{0^{2} }{2} + C\\1 = C[/tex]
[tex]y e^{t } = \frac{t^{2} }{2} + 1\\y(t) = \frac{1}{e^{t } } (\frac{t^{2} }{2} + 1)[/tex]
The value of y(1) will be expressed as;
[tex]y(1) = \frac{1}{e^{1} } (\frac{1^{2} }{2} + 1)\\y(1) = \frac{1}{2e}+\frac{1}{e}\\ y(1) = \frac{3}{2e}[/tex]
[tex]y(1) = \frac{3}{5.4366} \\y(1) = 0.5518[/tex]
