Answer:
The solution of the given initial value problems in explicit form is [tex]y=x-x^2-2[/tex] and the solutions are defined for all real numbers.
Step-by-step explanation:
The given differential equation is
[tex]y'=1-2x[/tex]
It can be written as
[tex]\frac{dy}{dx}=1-2x[/tex]
Use variable separable method to solve this differential equation.
[tex]dy=(1-2x)dx[/tex]
Integrate both the sides.
[tex]\int dy=\int (1-2x)dx[/tex]
[tex]y=x-2(\frac{x^2}{2})+C[/tex] [tex][\because \int x^n=\frac{x^{n+1}}{n+1}][/tex]
[tex]y=x-x^2+C[/tex] ... (1)
It is given that y(1) = -2. Substitute x=1 and y=-2 to find the value of C.
[tex]-2=1-(1)^2+C[/tex]
[tex]-2=1-1+C[/tex]
[tex]-2=C[/tex]
The value of C is -2. Substitute C=-2 in equation (1).
[tex]y=x-x^2-2[/tex]
Therefore the solution of the given initial value problems in explicit form is [tex]y=x-x^2-2[/tex] .
The solution is quadratic function, so it is defined for all real values.
Therefore the solutions are defined for all real numbers.
