Solve the Initial Value Problem and Plot the Particular Solution with the direction field (MAPLE)

Question

I am fairly new at MAPLE and I'm having some trouble solving this ODE.

$$(t+1)\frac{dy}{dt}-2(t^2+t)y=\frac{e^{t^2}}{t+1}$$

My initial value problem is $$t>-1, y(0)=5$$

I put the equation in standard form and typed into maple

However I am aware that when my initial value is $y(0)=5$ my equation becomes $4=0$ which is not possible. So I am confused on whether or not it is possible to find a general solution.

Answer 1

The image of your Maple code shows an italic e, which indicates that you may have tried to just type the letter "e" as the base of the natural logarithm. That would be committing a common usage mistake.

If entered correctly the terms like exp(t^2) would get displayed in 2D Output as an upright Roman "e" rather than with an italic "e" as the base.

In 1D plaintext Maple code the input would look like as follows (where I also show 1D output just for simplicity of my inlining the code here).

restart;

eq := diff(y(t),t)*(t+1) - 2*(t^2+t)*y(t) = exp(t^2)/(t+1);

                                                          2
               /d      \               2             exp(t )
         eq := |-- y(t)| (t + 1) - 2 (t  + t) y(t) = -------
               \dt     /                              t + 1

dsolve({eq, y(0)=5});

                      /    1      \      2
               y(t) = |- ----- + 6| exp(t )
                      \  t + 1    /

normal(%);

                                      2
                       (5 + 6 t) exp(t )
                y(t) = -----------------
                             t + 1

To enter exp(t^2) while in (default) fancy 2D Input mode in a Maple Document, you have three choices: 1) Just type it in as exp(t^2) 2) Use the Expression palette in the GUI's left panel which has an item for inserting a typeset 2D e^blah . 3) Type the letters exp and then the Esc key, to get a popup of command-completion suggestions, one of which is in the typeset e^blah form.

The keyboard acceleration to invoke command-completion varies by platform, and on Linux the Esc key does it. See the Help system for details, or look here.

Answer 2

Hint $$(t+1)\frac{dy}{dt}-2(t^2+t)y=\frac{e^{t^2}}{t+1}$$ Since $t >-1$ then $t \neq -1$ $$y'-2ty=\frac{e^{t^2}}{(t+1)^2}$$ $e^{-t^2}$as integrating factor $$(ye^{-t^2})'=\frac{1}{(t+1)^2}$$ $$(ye^{-t^2})=\int \frac{dt}{(t+1)^2}$$ $$y=e^{t^2}\int \frac{dt}{(t+1)^2}$$ Substitute $u=t+1$ $$y=e^{t^2}\int \frac{du}{u^2}$$ The general solution is $$y=e^{t^2}(K-\frac{1}{t+1})$$ $$y(0)=5 \implies K-1=5 \implies K=6$$ Solution with initial condition $$\boxed {y(t)=e^{t^2}\left(\frac{6t+5}{t+1}\right)}$$