y' = (1 − 2x) y^2

Divide differential equation with y2 to get classic separable equation. This is permissible for all y≠q0 Integrate both sides with respect to x Mind the chain rule and notice that integration of the left side can be written simply as integration with respect to y ∫1y2,dy=∫(1−2x),dx −1y+C1=x−x2+C2 1y=x2−x+C1−C2 Since C1 and C2 are constants C1−C2 is also a constant. Call it C General solution in implicit form is 1y=x2−x+C General solution in explicit form is 1x2−x+C Use initial condition to calculate constant C y(0)=−16102−0+C=−16C=−6 Therefore particular solution is Solution is not defined when x2−x−6=0, that is solution is not defined for x=−2,3 Solution is therefore well defined on intervals &lt;−∞,−2&gt;,&lt;−2,3&gt; and &lt;3,+∞&gt; Since solution curve must pass trough point (0,d−16) and 0∈&lt;−2,3&gt; we see that this solution is defined on interval &lt;−2,3&gt;

Resolved: "y' = (1 − 2x) y^2"

Yandel Hunter

Answered question

2022-02-10

y' = (1 − 2x) y^2

Answer & Explanation

sigcinweq0s

Beginner2022-02-11Added 19 answers

Divide differential equation with

y^{2}

to get classic separable equation.
This is permissible for all

y \neq q 0

Integrate both sides with respect to x
Mind the chain rule and notice that integration of the left side can be written simply as integration with respect to y

\int \frac{1}{y^{2}}, d y = \int (1 - 2 x), d x

\frac{- 1}{y} + C_{1} = x - x^{2} + C_{2}

\frac{1}{y} = x^{2} - x + C_{1} - C_{2}

Since

C_{1}

and

C_{2}

are constants

C_{1} - C_{2}

is also a constant.
Call it C
General solution in implicit form is

\frac{1}{y} = x^{2} - x + C

General solution in explicit form is

\frac{1}{x^{2} - x + C}

Use initial condition to calculate constant C

\begin{aligned} y (0) & = - \frac{1}{6} \\ \frac{1}{0^{2} - 0 + C} & = - \frac{1}{6} \\ C & = - 6 \end{aligned}

Therefore particular solution is
Solution is not defined when

x^{2} - x - 6 = 0

, that is solution is not defined for

x = - 2, 3

Solution is therefore well defined on intervals

< - \infty, - 2 >, < - 2, 3 >

and

< 3, + \infty >

Since solution curve must pass trough point

(0, d \frac{- 1}{6})

and

0 \in < - 2, 3 >

we see that this solution is defined on interval

< - 2, 3 >

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