Numerical Algorithms Test Suite

ODE Problem 1

IVP to solve:

$$\begin{array}{c}\frac{dy}{dt}=\frac{2y(t)-1}{t+3},y(0)=-1\end{array}$$

Analytical solution:

$$\begin{array}{rl}\frac{dy}{dt}=\frac{2y(t)-1}{t+3}& \Rightarrow \frac{dy}{2y(t)-1}=\frac{dt}{t+3}\\ & \Rightarrow \int \frac{dy}{2y(t)-1}=\int \frac{dt}{t+3}\\ & \Rightarrow \frac{1}{2}\ln |2y(t)-1|=\ln |t+3|+C\end{array}$$ $$\begin{array}{rl}\frac{1}{2}\ln |2y(t)-1|=\ln |t+3|+C& \Rightarrow \frac{1}{2}\ln |2(-1)-1|=\ln |0+3|+C\\ & \Rightarrow \frac{1}{2}\ln |-3|=\ln |3|+C\\ & \Rightarrow \frac{1}{2}\ln (3)=\ln (3)+C\\ & \Rightarrow -\frac{1}{2}\ln (3)=C\end{array}$$ $$\begin{array}{rl}\frac{1}{2}\ln |2y(t)-1|=\ln |t+3|-\frac{1}{2}\ln (3)& \\ & \Rightarrow \ln |2y(t)-1|=2\ln |t+3|-\ln (3)\\ & \Rightarrow \ln |2y(t)-1|=\ln |(t+3{)}^2|-\ln (3)\\ & \Rightarrow \ln |2y(t)-1|=\ln |\frac{(t+3{)}^2}{3}|\\ & \Rightarrow 2y(t)-1=\frac{(t+3{)}^2}{3}\\ & \Rightarrow y(t)=\frac{(t+3{)}^2}{6}-\frac{1}{2}\end{array}$$

ODE Problem 2

IVP to solve:

$$\begin{array}{c}(1+t^2)y^{\prime }(t)-2ty(t)=t,y(0)=0\end{array}$$

Analytical solution (with steps):

$$\begin{array}{rl}(1+t^2)y^{\prime }(t)-2ty(t)=t& \Rightarrow y^{\prime }(t)-\frac{2t}{1+t^2}y(t)=\frac{t}{1+t^2}\end{array}$$ $$\begin{array}{c}p(t)=-\frac{2t}{1+t^2},q(t)=\frac{t}{1+t^2}\end{array}$$ $$\begin{array}{rl}\mu & =\exp (\int p(t)dt)\\ & =\exp (\int -\frac{2t}{1+t^2}dt)\\ & =\exp (-\ln (1+t^2))\\ & =\exp (\ln (\frac{1}{1+t^2}))\\ & =\frac{1}{1+t^2}\end{array}$$ $$\begin{array}{rl}y(t)& =\frac{1}{\mu }[\int \mu q(t)dt+C]\\ & =(1+t^2)[\int \frac{1}{(1+t^2{)}^2}dt+C]\\ & =(1+t^2)[\frac{1}{2(1+t^2)}\ln |(1+t^2{)}^2|+C]\\ & =\frac{1}{2}\ln |(1+t^2)|+C(1+t^2)\end{array}$$ $$\begin{array}{c}y(t)=\frac{1}{2}\ln |(1+t^2)|\end{array}$$