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never [62]
3 years ago
13

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
Mathematics
1 answer:
natali 33 [55]3 years ago
4 0

Answer:

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.

Step-by-step explanation:

The given differential equation is

y'=1-2x

It can be written as

\frac{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(\frac{x^2}{2})+C                  [\because \int x^n=\frac{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.

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