Question

Find the solution of the given initial value problems in explicit form. Determine the interval where the solutions are defined. y' = 1-2x, y(1) = -2

Answer 1

The solution of the given initial value problems in explicit form is
`y=x-x^2-2` and the solutions are defined for all real numbers.

Explanation:

The given differential equation is

`y'=1-2x`

It can be written as

`(dy)/(dx)=1-2x`

Use variable separable method to solve this differential equation.

`dy=(1-2x)dx`

Integrate both the sides.

`\int dy=\int (1-2x)dx`

`y=x-2((x^2)/(2))+C`
`[\because \int x^n=(x^(n+1))/(n+1)]`

`y=x-x^2+C` ... (1)

It is given that y(1) = -2. Substitute x=1 and y=-2 to find the value of C.

`-2=1-(1)^2+C`

`-2=1-1+C`

`-2=C`

The value of C is -2. Substitute C=-2 in equation (1).

`y=x-x^2-2`

Therefore the solution of the given initial value problems in explicit form is
`y=x-x^2-2` .

The solution is quadratic function, so it is defined for all real values.

Therefore the solutions are defined for all real numbers.