Consider an ordinary differential equation of the form : dy/dt =2t+3 ,
Now we take the formula for dy/dt and cross XXXXXXXX so that all the t's XXX XX one side XXX XXX y's on XXXXX.
XX XXXX XX have,
dy= (2t +X) dt
XXX we XXXX XX integrate XXXX XXXXX.
XXXXX integration we XXX XXX solution XX,
y=t^2 + 3t + X , XXXXX C is XXX XXXXXXXX of XXXXXXXXXXX.
XXXXXX one XX Integrating XXXXXX Method:
Here we XXXX to XXXXX XXXXX the XXXXXXXXXXXX equation in XXX form: dy/dt+p(t)y=X(t)
Consider an example,
dy/dt+2y=3
XXXX p(t)=2 and q(t) =X
XXX XX XXXX to XXXX an XXXXXXXXXXX XXXXXX i.e. u(t)=e^(XXXXXXXX of p(t) dt)
i.e. e^{Integral 2dt}= e^(2t)
XXX we have XX multiply XXX whole XXXXXXXXXXXX XXXXXXXX with the integrating factor. i.e.
e^(2t) dy/dt+XX^(XX) y=XX^(2t) -----> (1)
XXX the XXXX XXXX XXXX XXX be written as a XXXXX XXXXXXXXXX i.e.
e^(2t)dy/dt+XX^(2t)y=(e^(2t)y)'
Substituting XXXX in (X) we get,
(e^(XX)y)'=XX^(2t)
Now integrating both sides we get,
e^(2t)y=e^(XX)+C
i.e.
y=X+XX^(-2t) XX XXX solution with XXXXXXXX X.