Exercise
$t^2\cdot\frac{dx}{dt}+5tx=t^4\cdot ln\left(t\right)+5$
Step-by-step Solution
Learn how to solve problems step by step online. Solve the differential equation t^2dx/dt+5tx=t^4ln(t)+5. Divide all the terms of the differential equation by t^2. Simplifying. We can identify that the differential equation has the form: \frac{dy}{dx} + P(x)\cdot y(x) = Q(x), so we can classify it as a linear first order differential equation, where P(t)=\frac{5}{t} and Q(t)=\frac{t^4\ln\left(t\right)+5}{t^2}. In order to solve the differential equation, the first step is to find the integrating factor \mu(x). To find \mu(t), we first need to calculate \int P(t)dt.
Solve the differential equation t^2dx/dt+5tx=t^4ln(t)+5
Final answer to the exercise
$x=\frac{1}{t^{5}}\left(\frac{t^{8}\ln\left(t\right)+5t^{4}}{4}+\frac{-8t^{8}\ln\left(t\right)-t^{8}}{64}+C_0\right)$