# Math Help - DfEq: solution that passes through a point

1. ## DfEq: solution that passes through a point

Hello,
I have this diffyQ problem that asks to find the solution that passes through a specific point. I have worked it out to where I have a ( i think correct ) equation for y, but am not sure how to deal with the constant and make sure it passes through a specific point.

Solve: Find the solution of y' = 2(2x-y) that passes through the point P(0,1).
Solution:
y' = 4x -2y
y' + 2y = 4x
integrating factor is e^(2x)
d/dx of [e^(2x)y] = 4xe^(2x)
integrate both sides
RHS by parts gives
2xe^(2x) - e^2x
So:
e^(2x)y = 2xe^(2x) - e^(2x) + C
y = 2x - 1 + Ce^(-2x)

Could someone check this out and tell me how to get a solution that passes through that point? Thanks!

2. Hello, pberardi!

You may kick yourself when you see how easy it is . . .

Find the solution of $y' \:= \:2(2x-y)$ that passes through the point $P(0,1).$

Solution: . $y \:= \:2x - 1 + Ce^{-2x}$

$\text{"Passes through P(0,1)" means that: }\:x = 0,\:y = 1\,\text{ satisfies the equation.}$

$\text{Plug them in: }\;1 \;=\;2(0) - 1 + Ce^0 \quad\Rightarrow\quad 1 \:=\:\text{-}1 + C \quad\Rightarrow\quad C \:=\:2$

Solution: . $y \;=\;2x - 1 + 2e^{-2x}$

3. Thank you Soroban

4. Hi.

Originally Posted by pberardi
Hello,
I have this diffyQ problem that asks to find the solution that passes through a specific point. I have worked it out to where I have a ( i think correct ) equation for y, but am not sure how to deal with the constant and make sure it passes through a specific point.

Solve: Find the solution of y' = 2(2x-y) that passes through the point P(0,1).
Solution:
y' = 4x -2y
y' + 2y = 4x
integrating factor is e^(2x)
d/dx of [e^(2x)y] = 4xe^(2x)
integrate both sides
RHS by parts gives
2xe^(2x) - e^2x
So:
e^(2x)y = 2xe^(2x) - e^(2x) + C
y = 2x - 1 + Ce^(-2x)

Could someone check this out
Yes, it is correct.
You just need to verify that

y' = 2 - 2*C*e^(- 2x) = 2(2x-y)

and tell me how to get a solution that passes through that point? Thanks!
f(x):=y=2x - 1 + Ce^(-2x)

P(0,1)

f(0) = 2*0 - 1 + C *e^(-2*0) = 1

<=> -1 + C*1 = 1

<=> -1 + C = 1

Find C :-)

Regards, Rapha

Edit: Darn, beaten by 2 minutes. What ever