Question

Calculus2 first order differential equation y' + (4y/(100+t)) = 10. Can i get help starting this problem. Thanks. :)

Answer 1

You have a linear ODE, so you can start by finding the integrating factor.
\[\mu(t)=\exp\left(\int\frac{4}{100+t}\,\mathrm dt\right)=\exp\left(4\ln(100+t)\right)=(100+t)^4\]Multiply both sides of the ODE by \(\mu(t)\) and you get
\[(100+t)^4y'+4(100+t)^3y=10(100+t)^4\]Now the left hand side can be consolidated as the derivative of a product. Because
\[\frac{\mathrm d}{\mathrm dt}\bigg[(100+t)^4\bigg]=4(100+t)^3\]you have
\[(100+t)^4y'+4(100+t)^3y=\frac{\mathrm d}{\mathrm dt}\bigg[(100+t)^4y\bigg]\]So for the remaining form of the ODE,
\[\frac{\mathrm d}{\mathrm dt}\bigg[(100+t)^4y\bigg]=10(100+t)^4\]you can integrate both sides with respect to \(t\) to get
\[\int\frac{\mathrm d}{\mathrm dt}\bigg[(100+t)^4y\bigg]\,\mathrm dt=\color{red}{(100+t^4)y=\int10(100+t)^4\,\mathrm dt}\]It's easy to solve for \(y\) from here.