When a derivative can be written as a function of the independent variable only for example y'=f(x) y''=f(x) y'''(x)=f(x) etc. We call that a directly integrable equation. Example 1: Find the general solution for the following directly integrable equation ![diff(y, x) = 6*x^2+4*y(1) and 6*x^2+4*y(1) = 0]()
That means ![int(6*x^2+4, x)]()
![y = 2*x^3+c+4*x]() c is an arbitary solution
![``]()
![equation1 := diff(y(x), x) = 6*x^2+4]() ![diff(y(x), x) = 6*x^2+4]()
| (1) |
![NULL]()
![NULL]()
![sol1 := dsolve(equation1, y(x))]()
![y(x) = 2*x^3+c__1+4*x]()
| (2) |
And if we have the initial condition
![y(1) = 0]()
![particular_sol1 := dsolve({equation1, y(1) = 0}, y(x))]() ![y(x) = 2*x^3+4*x-6]()
| (3) |
!["(->)"]()
!["Example 2:"]() ![NULL]()
![NULL]()
![" Find the particular solution for the following equation with condition"]()
![x^2*(diff(y(x), x)) = -1]()
y(1)=3 So we will need to get the y' by itself ![int(-1/x^2, x)]()
so, , where c is an arbitary constant
And this is our general solution. Now we plug in the initial condition when . That means ![c = 2.]() Thus, the particular solution is ![y = 1/x+2]() ![``]()
![eq := x^2*(diff(y(x), x)) = -1]()
![x^2*(diff(y(x), x)) = -1]()
| (4) |
![NULL]()
![NULL]()
![sol := dsolve(eq, y(x))]()
![y(x) = 1/x+c__1]()
| (5) |
![NULL]()
![particular_sol := dsolve({eq, y(1) = 3}, y(x))]()
![y(x) = 1/x+2]()
| (6) |
![NULL]()
![NULL]()
![plot(1/x+2, x = -20 .. 20, color = "Red", axes = normal, legend = [typeset(1/x+2)])]()
![NULL]()
![NULL]()
![NULL]()
![NULL]()
![" Example 3:"]()
![" Find the particular solution for the following equation with condition"]()
![diff(y, t, t) = cost, (D(y))(0) = 0, y(0) = 1]()
![eq1 := diff(y(t), t, t) = cos(t)]()
![diff(diff(y(t), t), t) = cos(t)]()
| (7) |
![particular_sol := dsolve({eq1, y(0) = 1, (D(y))(0) = 0}, y(t))]()
![y(t) = -cos(t)+2]()
| (8) |
!["(->)"]()
![NULL]()
|