Question

Solve the initial value problem
Y (prime)+2xy=×, y (0)=2

Answer 1

\[y'+2x \cdot y=x\]
I would first finding the integrating factor \[v=e^{ \int\limits 2x dx}\]

Answer 2

\[v y'+v 2x \cdot y=v x \\ (v y)'=vx \]
Find v and plug it in
then integrate both sides of (vy)'=vx with respect to x

Answer 3

V=e^x^2. Right?

Answer 4

\[(e^{x^2}y)' =e^{x^2}x\]

Answer 5

yep now integrate both sides

Answer 6

You you only multiplying y? Don't we have to multiply the whole equation?

Answer 7

Why*

Answer 8

I'm sorry. What?

Answer 9

I think I did multiply the whole equation by v as shown above

Answer 10

When you integrate both sides you get
e^x^2 y= e^x^2+ C. Then divide both sides by e^x^2 which gives you y= 1+C e^-×^2. Then y (0)=2. We get C=2. Is it correct?

Answer 11

\[e^{x^2} y =\int x e^{x^2} dx \text{ let } u=x^2 \text{ so } du=2x dx \text{ } \frac{1}{2} du=x dx \\ e^{x^2}y=\int\limits_{}^{} \frac{1}{2} e^{u} du =\frac{1}{2}e^u+C =\frac{1}{2}e^{x^2}+C\] \\ \[e^{x^2}y=\frac{1}{2}e^{x^2}+C \\ y=\frac{1}{2}+Ce^{-x^2}\]

Answer 12

now if you have y(0)=2 then replace x's with 0 and y's with 2 and solve for C

Answer 13

\[2=\frac{1}{2}+C\]

Answer 14

Thank you very much

Answer 15

np