Question

Solve the following initial value problem:

t* dy/dt+3y=8t
with y(1)=8.

Find the integrating factor, u(t)= ,
and then find y(t)=

Answer 1

We are asked the solve the following differential equation,
`t(dy)/(dt)+3y=8t`. This is an ordinary, linear, first-order DE. We need it in the form
`(dy)/(dt)+P(t)y=Q(t)`, where
`\mu(t)=e^{\int\ {P(t)} \, dt }`, which is the integrating factor.

`\Longrightarrow t(dy)/(dt)+3y=8t`, multiply everything by t.

`\Longrightarrow (dy)/(dt)+((3)/(t) )y=8`

We now have the correct form for solving a linear, first-order DE.

Find
`\mu(t)`.

`\Longrightarrow \mu(t)=e^{\int\ {P(t)} \, dt } \Longrightarrow \mu(t)=e^{\int\ {(3)/(t) } \, dt }`

`\Longrightarrow \mu(t)=e^{3\int\ {(1)/(t) } \, dt }\Longrightarrow \mu(t)=e^(3ln|t|) \Longrightarrow \mu(t)=e^(ln(t^3))`

`\Longrightarrow \mu(t)=t^3`

Now multiply the entire DE by
`\mu(t)`.

`\Longrightarrow t^3[(dy)/(dt)+((3)/(t) )y=8]`

`\Longrightarrow (t^3)(dy)/(dt)+(t^3)((3)/(t) )y=(t^3)8`

`\Longrightarrow (t^3)(dy)/(dt)+(3t^2)y=(t^3)8`

Verify that the L.H.S is a product rule of
`t^3y`.

Product rule:
`(d)/(dx)[f(x)g(x)]=f(x)g'(x)+f'(x)g(x)`

`(d)/(dt)[t^3y]=(t^3)((dy)/(dt) )+(3t^2)(y)`

It is the product rule, so we can move forward by integrating both sides of the DE.

`\Longrightarrow \int\ {[t^3y]'} \, =\int\ {8t^3} \, dt`

`\Longrightarrow t^3y=2t^4+C`

Now using the initial condition to find the arbitrary constant, C.

`t^3y=2t^4+C ;y(1)=8`

`\Longrightarrow (1)^3(8)=2(1)^4+C`

`\Longrightarrow 8=2+C`

`\Longrightarrow 6=C \Longrightarrow C=6`

Now we have,
`t^3y=2t^4+6`, solve for y.

`\Longrightarrow t^3y=2t^4+6`

`\Longrightarrow y=2(t^4)/(t^3) +(6)/(t^3)`

`\Longrightarrow y=2t +6t^(-3)`

Thus the given differential equation with the initial condition is solved,
`y=2t +6t^(-3)`.

Answer 2

The integrating factor, u(t), for the initial value problem t* dy/dt+3y=8t with y(1)=8 is u(t) = e^(3t). The solution to this equation is y(t) = 8e^(-3t)+2te^(-3t).

Explanation:

The integrating factor, u(t), for the given initial value problem t(dy/dt)+4y=3t with y(1)=8 is u(t) = e^(4∫t dt) = e^(2t^2).

Using this integrating factor, the solution to the initial value problem is y(t) = (3/8)te^(2t^2)+Ce^(-2t^2), where C is a constant of integration. Since y(1)=8, we can solve for C and find that C=-7/4. Therefore, the solution to the initial value problem is y(t) = (3/8)te^(2t^2)-7/4e^(-2t^2).