Question

solve the initial value problem.
$t^2 \frac{dx}{dt} + 4tx = t^6 \ln t + 2$, $x(1) = 0$

the solution is $x(t) = \square$.
(use parentheses to clearly denote the argument of each function.)

Explanation:

Step1: Rewrite as linear ODE

Divide the equation \( t^{2}\frac{dx}{dt}+4tx = t^{6}\ln t + 2 \) by \( t^{2} \) to get the standard linear first - order ODE form \(\frac{dx}{dt}+P(t)x = Q(t)\):
\(\frac{dx}{dt}+\frac{4}{t}x=t^{4}\ln t+\frac{2}{t^{2}}\)
Here, \( P(t)=\frac{4}{t} \) and \( Q(t)=t^{4}\ln t+\frac{2}{t^{2}} \)

Step2: Find the integrating factor

The formula for the integrating factor \(\mu(t)\) of a linear ODE \(\frac{dx}{dt}+P(t)x = Q(t)\) is \(\mu(t)=e^{\int P(t)dt}\)
Calculate \(\int P(t)dt=\int\frac{4}{t}dt = 4\ln|t|=\ln(t^{4})\) (assuming \(t>0\) since we have \(\ln t\) in the equation)
Then \(\mu(t)=e^{\ln(t^{4})}=t^{4}\)

Step3: Multiply the ODE by the integrating factor

Multiply \(\frac{dx}{dt}+\frac{4}{t}x=t^{4}\ln t+\frac{2}{t^{2}}\) by \(t^{4}\):
\(t^{4}\frac{dx}{dt}+4t^{3}x=t^{8}\ln t + 2t^{2}\)
Notice that the left - hand side is the derivative of the product \(\mu(t)x\) by the product rule \((uv)^\prime = u^\prime v+uv^\prime\) where \(u = t^{4}\) and \(v = x\). So, \(\frac{d}{dt}(t^{4}x)=t^{8}\ln t + 2t^{2}\)

Step4: Integrate both sides

Integrate the left - hand side with respect to \(t\): \(\int\frac{d}{dt}(t^{4}x)dt=t^{4}x\)
Integrate the right - hand side:
First, integrate \(\int t^{8}\ln tdt\) using integration by parts. Let \(u=\ln t\), \(dv = t^{8}dt\)
Then \(du=\frac{1}{t}dt\), \(v=\frac{t^{9}}{9}\)
By the integration by parts formula \(\int u\;dv=uv-\int v\;du\), we have:
\(\int t^{8}\ln tdt=\frac{t^{9}}{9}\ln t-\int\frac{t^{9}}{9}\cdot\frac{1}{t}dt=\frac{t^{9}}{9}\ln t-\frac{1}{9}\int t^{8}dt=\frac{t^{9}}{9}\ln t-\frac{t^{9}}{81}+C_1\)
Second, integrate \(\int 2t^{2}dt=\frac{2t^{3}}{3}+C_2\)
So, \(\int(t^{8}\ln t + 2t^{2})dt=\frac{t^{9}}{9}\ln t-\frac{t^{9}}{81}+\frac{2t^{3}}{3}+C\)

Step5: Solve for \(x\)

We have \(t^{4}x=\frac{t^{9}}{9}\ln t-\frac{t^{9}}{81}+\frac{2t^{3}}{3}+C\)
Divide both sides by \(t^{4}\):
\(x(t)=\frac{t^{5}}{9}\ln t-\frac{t^{5}}{81}+\frac{2}{3t}+ \frac{C}{t^{4}}\)

Step6: Apply the initial condition \(x(1) = 0\)

Substitute \(t = 1\) and \(x(1)=0\) into the equation for \(x(t)\):
\(0=\frac{1^{5}}{9}\ln(1)-\frac{1^{5}}{81}+\frac{2}{3\times1}+\frac{C}{1^{4}}\)
Since \(\ln(1) = 0\), the equation simplifies to:
\(0=0-\frac{1}{81}+\frac{2}{3}+C\)
\(C=\frac{1}{81}-\frac{2}{3}=\frac{1 - 54}{81}=-\frac{53}{81}\)

Answer:

\(x(t)=\frac{t^{5}}{9}\ln(t)-\frac{t^{5}}{81}+\frac{2}{3t}-\frac{53}{81t^{4}}\)