Question

Solve the following initial value problem: dy +9y 6 dt with y(1) = 8 у:

Answer 1

Answer:

The solution to the initial value problem is $y(t)=\frac{3t^{10}+37}{5t^9}$

Step-by-step explanation:

We have the following differential equation

$t\frac{dy}{dt}+9y=6t, \:{y(1)=8}$

This is a linear differential equation because we can made to look like this:

$\frac{dy}{dt}+P(t)y=Q(t)$

where $P(t)$ and $Q(t)$ are functions of t.

The solution process for a first-order linear differential equation is as follows.

1. Put the differential equation in the correct initial form.

2. Find the integrating factor, $\mu(t)$.

3. Multiply everything in the differential equation by $\mu(t)$ and verify that the left side becomes the product rule $(\mu(t)y(t))'$ and write it as such.

4. Integrate both sides, make sure you properly deal with the constant of integration.

5. Solve for the solution $y(t)$.

Applying the solution process, we have

1. We need to get the differential equation in the correct form.

$\frac{dy}{dt}+\frac{9}{t} y=6$

From this we can see that $p(t)=\frac{9}{t}$

2. The integrating factor is $\mu(t) = e^{\int {p(t)} \, dt }$

$\mu(t) = e^{\int {p(t)} \, dt }\\\mu(t) = e^{\int {\frac{9}{t} } \, dt }\\\mu(t) = e^{9\ln \left|t\right|}\\\mu(t) = t^9$

3. Now multiply all the terms in the differential equation by the integrating factor and do some simplification.

$\frac{dy}{dt}+\frac{9}{t} y=6\\t^9(\frac{dy}{dt})+t^9(\frac{9}{t} y)=6t^9\\(t^9y)'=6t^9$

4. Integrate both sides and don't forget the constants of integration that will arise from both integrals.

$\int {(t^9y)'} \, dt =\int {6t^9} \, dt \\\\t^9 y+k=\frac{3t^{10}}{5}+c\\\\t^9 y=\frac{3t^{10}}{5}+c-k\\\\t^9 y=\frac{3t^{10}}{5}+d\\\\y=\frac{3t^{10}+5d}{5t^9}$

5. To find the solution we are after we need to identify the value of d that will give us the solution we are after. To do this we simply plug in the initial condition which will give us an equation we can solve for d.

$8=\frac{3(1)^{10}+5d}{5(1)^9} \Rightarrow d=\frac{37}{5}$

So, the actual solution to the initial value problem is

$y(t)=\frac{3t^{10}+37}{5t^9}$