Question

Consider the differential equation:

Answer 1

Answer:

(a) First order linear separable differential equation

(b)

$y(x)=2+C_1e^{x-x^2} \\\\I=(-\infty,\infty)$

(c)

(d)

$y(x)=2-e^{x-x^2}$

Step-by-step explanation:

(b) Solve for $\frac{dy}{dx}$

$\frac{dy}{dx}=4x+y-2xy-2\\ \\Simplify\\\\\frac{dy}{dx}=(2x-1)(2-y)\\\\Divide\hspace{3}both\hspace{3}sides\hspace{3}by\hspace{3}2-y\hspace{3}and\hspace{3}multiply\hspace{3}both\hspace{3}sides\hspace{3}by\hspace{3}dx\\\\\frac{dy}{2-y}=(2x-1) dx\\\\Integrate\hspace{3}both\hspace{3}sides\\\\\int\frac{dy}{2-y} \, =\int\ (2x-1) dx\\\\Evaluate\hspace{3}the\hspace{3}integrals\\\\-log(2-y)=x^2-x+C_1\\\\Solving\hspace{3}for\hspace{3}y\\\\y=2+C_1e^{x-x^2}$

The domain of y is: $x\in R\hspace{3}or\hspace{3}(-\infty,\infty)$

So the lasrgest interval I on which the solution is defined is:

$I=(-\infty,\infty)$

(c)

Differentiate y:

$\frac{dy}{dx}=C_1(1-2x)e^{x-x^2} =C_1e^{x-x^2}-2xC_1e^{x-x^2}$

Evaluate this result into the differential equation:

$\frac{dy}{dx}=4x+y-2xy-2\\\\C_1e^{x-x^2} -2xC_1e^{x-x^2} =4x+2+C_1e^{x-x^2}-4x-2xC_1e^{x-x^2}-2\\\\C_1e^{x-x^2} -2xC_1e^{x-x^2} =C_1e^{x-x^2} -2xC_1e^{x-x^2}$

Therefore, the solution is correct.

(d)

Simply evaluate the function y for x=0 and solve for C1:

$y(0)=2+C_1e^{0-0^2} =1\\\\2+C_1e^0=1\\\\2+C_1*1=1\\\\2+C_1=1\\\\C_1=-1$

Finally substitute into y:

$y(x)=2-e^{x-x^2}$