Question

how to solve the Differential equation: y'+2xy=2x , y(0)=2 how to solve the Differential equation: y'+2xy=2x , y(0)=2 @Mathematics

Answer 1

do you know Laplace Transformation?

Answer 2

turn y' into dy/dx first

Answer 3

no, im not sure how to do the laplace transformation.

Answer 4

ok..
dy/dx-2xy=2x (1)
The differential equation without the second member is the following.
dy/dx-2xy=0
dy/dx=2xy
dy/y=2xdx
ln|y|=x²+lnC
y=Ce^x²

Let's note dy/dx=y'.

We can use the method of the variation of the constant.
y'=C'e^x²+2xCe^x²
(1) is written as follows.
C'e^x²+2xCe^x²-2xCe^x²=2x
C'e^x²=2x
C'=2xe^-x²

C=-e^-x²+K

y=Ce^x²
y=-1+Ke^x²

The solution of the differential equation is
y=Ke^x²-1

Answer 5

Since this is in the form y' + p(x)y=g(x), it is a linear differential equation and can be solved using the integrating factor \[u(x)=e ^{\int\limits_{}^{}p(x)dx}\]

Answer 6

Since p(x)=2x, the integrating factor is \[u(x)=e ^{x ^{2}}\]

Answer 7

Multiplying by the integrating factor, our equation changes to\[e ^{x ^{2}}y' + 2xe ^{x ^{2}}y = 2xe ^{x ^{2}}\]

Answer 8

We can take the integral of both sides. Notice that the left side looks like what occurs when we take the derivative using the product rule, so when we integrate that side we end up with a product.
\[e ^{x ^{2}}y=e^{x ^{2}}+C\]

Answer 9

Divide to get y by itself and we have\[y=1+Ce^{-x ^{2}}\]

Answer 10

Using the initial conditions of y=2 when x=0, we can solve for C and get C=1.

Answer 11

thank you! :)

Answer 12

You're welcome.

Answer 13

y'+2xy=2x , y(0)=2
dy/dx+2xy-2x =0
dy+2xydx-2xdx =0
dy+2xdx(y-1) =0
dy=-2xdx(y-1)
dy/(y-1) =-2xdx integrating both sides gives

Answer 14

ln(y-1)=-x^2 +c
whenx=0,y=2
c=0
y-1=e^(-x^2)
y=1+e^(-x^2) ans

Answer 15

this is the solution to the DE y'+2xy=2x , y(0)=2

Answer 16

You did it by separating them. I didn't see that. :-)