Question

Verify that the function is a solution of the differential equation on some interval, for any choice of the arbitrary constants appearing in the function.
(a) y = c e^(2x) ; y' =2y

(b) y = (x^2)/3 + c/x ; xy' + y = x^2

(c) y = 1/2 + c e^((-x^2)/2) ; y' + 2xy = x

(d) y = (1+c e^((-x^2)/2)) ; y = (1-c e^((-x^2)/2) ; 2y' + x(y^2 -1) = 0

Answer 1

Answer:

See details below

Step-by-step explanation:

To verify that y=f(x) is a solution of the differential equation in some interval (where f is defined), we take derivatives respect to x using the usual rules (sum, product, chain rule, exponential,...)

a) $y=ce^{2x}\rightarrow y'=(2x)'ce^{2x}=2ce^{2x}=2y$

b) $y=(x^2)/3+c/x \rightarrow xy'+y=x( 2x/3 -c/x^{2} )+(x^2)/3+c/x =3x^2/3-c/x+c/x=x^2$

c) The solution of the differential equation is slightly different (the function given is a solution only if c=0)

$y=1/2 + ce^{(-x^2)} \rightarrow y'+2xy=((-x^2))'ce^{(-x^2)}+2x(1/2 + ce^{(-x^2)} )=-2cxe^{(-x^2)}+x+2cxe^{(-x^2)}=x$

d) $y=1+c e^{(-x^2)/2)} \rightarrow 2y'+x(y^2-1)=2(-xce^{x^2/2})+x(1+2ce^{x^2/2}+c^2e^{-x^2}-1)=-2xce^{x^2/2}+2xce^{x^2/2}+c^2e^{-x^2}=c^2e^{-x^2}=0$

only if c=0, this also happens for the other function.