Given $Re\{f'(z)\}$, to find $Im\{f(z)\}$ 
An analytic function $f(z)$ is such that $\Re\{f'(z)\} =2y$ and $f(1+i)=2$. Then the imaginary part of $f(z)$ is

*

*$-2xy$


*$x^2-y^2$


*$2xy$


*$y^2-x^2$

Here, by using Milne Thomson's method I get $f'(z)=\int{-2i}$ $dz$ $=-2iz+c$, $c$ is the ingrating constant.
Now again integrating I got $f(z) =-iz^2+cz+d$, $d$ is another integrating constant.
But by only one condition I can not find out the value of two constants $c$ and $d$. But if I omit $c$, then using the given condition I get $\Im(f(z))$ as $y^2-x^2$.
I'm really confused and cannot understand what to do actually. Should I omit $c$ in the first integration?
Is there any other method for finding the solution?
Update:(21st May-2015)
Using Cauchy-Riemann equations we get $f'(z)=2y-2ix$. Then ,
$$f(z)=\int (2y-x)(\,dx+i\,dy)+C$$
$$=\int 2y\,dx+2i\int y\,dy-2i\int x\,dx+2\int x\,dy+C$$
$$=2xy+iy^2-ix^2+2xy+C$$
$$=4xy+i(y^2-x^2)+C.$$
So, if we neglect the constant or take it as a real constant then we get the imaginary part is $y^2-x^2$.

But I have a confusion about the integration. Is the integration correct ? If wrong then please tell me why it is incorrect ?

There is no difficulty about the answer of  Daniel Fischer. See my update only.
 A: You are right, the data aren't sufficient to uniquely determine $f$.
Since we are given $\operatorname{Re} f'(z)$, we know that $c$ is purely imaginary, but it can be any purely imaginary constant. Before we take $f(1+i) = 2$ into account, we write the candidate for $f$ in the form
$$f(z) = -iz^2 + c(z - 1-i) + \tilde{d},$$
at which point $f(1+i) = 2$ reduces to $\tilde{d} = 0$, and we see that $\operatorname{Im} f(z)$ can be
$$y^2 - x^2 + r(x-1)$$
with an arbitrary $r\in\mathbb{R}$.
Among the four given choices, there is however only one that is possible.
A: $f^{'}(z) = 2y + iv(x,y)$
$u_x=v_y$ and $v_x = - u_y$
$0 = v_y $ and $2 = -v_x$
$v_x = -2$
Integrating with respect to $'x'$
$v = -2x + \phi(y)$
differentiating with respect to $'y'$
$v_y = 0 + \phi^{'}(y)$
therefore $\phi^{'}(y) = v_y$
But we know that $v_y = 0$, hence $\phi^{'}(y) = 0$.
therefore $v = -2x + c$.
Hence $f'(z)=2y + i(-2x+c) $
Using c-r equations
$f'(z) = a_x + i b_x$
$a_x = 2y$ and $b_x = -2x + c$
we need to find $Im(f)$ (i.e) $b(x,y)$
Method1:
so, integrating with respect to $x$,
$b = -x^2 + xc+ \chi(y)$
Now, $f(x+iy) = a+ib$ ,$f(1+i) = 2 + i0$
so,$ 0 = -(1)^2 + (1)(c) + \chi(y)$
$\chi(y) = c+1 $
Substitute, $b= xc - x^2 + c+1$
hence $im(f) = xc - x^2 + c+1$.
method 2: instead of finding $b(x,y)$ find $a(x,y)$
now $a_x = 2y$
integrating with respect to $'x'$, we get
$a = 2xy + g(y); f(x+iy) = a + ib$ (i.e) $f(1+i) = 2+i0$
therefore when $x=1,y=1$ implies $a=2$ and $b=0$
$2 = 2(1)(1) + g(y)$
$g(y) = 0; a = 2xy;$
now $f(z) = 2xy + i b(x,y)$
$a_x = 2y$ and $a_y = 2x$
By C-R equations, $a_x = b_y$ and $b_x = - a_y$
$b_y =2y$; integrating
$b=y^2 + h(x)$
Differentiating
$b_x = h^{'}(x)$
$-2x = h^{'}(x)$
$h(x) = -x^{2}$
so $b = y^2 - x^2$..
Hence the answer...
If there is a mistake, please let me know it.
A: Let $f'(z)=\frac{\partial u}{\partial x}+i \frac{\partial v}{\partial x}$
Then, it is given that $\frac {\partial u}{\partial x}=2y$ 
Integrating this we get, 
$u=2xy+\lambda(y)$ Where $\lambda (y)$ is an unknown function of $y$ that might have been lost in partial.
Calculating $\frac{\partial u}{\partial y}=2x+\lambda'(y)$
From CR equations, $\frac{\partial v}{\partial x}=-\frac{\partial u}{\partial y}$
It all gives us $f'(z)=2y-i(2x+\lambda'(y))$
If $z=x+iy$ then as $y\to 0$$\implies$ $x\to z$
$f'(z)=-i(2z+\lambda'(0))$
Integrating this we will get, 
$f(z)=-iz^2-iz\lambda'(0)+c_1$
Now using the given condition ,
$f(1+i)=2-i\lambda'(0)+\lambda'(0)+c_1$
$2=2-i\lambda'(0)+\lambda'(0)+c_1$
$0=c_1+\lambda'(0)-i\lambda'(0)$
Comparing real and imaginary part and assuming $c_1$ is real.
$\lambda'(0)=0$ and $c_1=0$
Gives:  
$f(z)=iz^2$
$f(z)=2xy+i(y^2-x^2)$
