Solving a Diff. Equation: y'=x+y, y(0)=2

[skyturnred]

Homework Statement

Solve:

y'=x+y, y(0)=2

Homework Equations

The Attempt at a Solution

I THINK my method is correct.. but I messed up somewhere.

I rearrange for y'-y=x

integrate both sides gets me:

y-$\frac{y^{2}}{2}$=$\frac{x^{2}}{2}$

after completing the square I get

(y-1)$^{2}$=-x$^{2}$+1

But this is where I mess up. To solve for y, I square root each side. But then I get '+ or -' on the right side:

y=$\pm$$\sqrt{-x^{2}+1}$+1+c

so solving for both cases gets me c=0 OR c=2. But I can only have once answer. Where did I go wrong?

 

[Ray Vickson]

Homework Statement

Solve:

y'=x+y, y(0)=2

Homework Equations

The Attempt at a Solution

I THINK my method is correct.. but I messed up somewhere.

I rearrange for y'-y=x

integrate both sides gets me:

y-$\frac{y^{2}}{2}$=$\frac{x^{2}}{2}$

after completing the square I get

(y-1)$^{2}$=-x$^{2}$+1

But this is where I mess up. To solve for y, I square root each side. But then I get '+ or -' on the right side:

y=$\pm$$\sqrt{-x^{2}+1}$+1+c

so solving for both cases gets me c=0 OR c=2. But I can only have once answer. Where did I go wrong?

Going from y'-y=x to $y-\frac{y^2}{2} =\frac{x^2}{2}$ is wrong. The DE reads as $dy - y dx = x dx,$ so when you integrate on the left you don't get y - y^2/2. You need to use an "integrating factor"; see http://www.ucl.ac.uk/Mathematics/geomath/level2/deqn/de8.html , or read your textbook.

RGV

 

[skyturnred]

Thanks! Your hint helped me get the right answer!