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A function $y(x)$ that satisfies $\dfrac{dy}{dx}+4xy=x$ with the boundary condition $y(0)=0$ is

  1. $y(x)=(1-e^x)$
  2. $y(x)=\frac{1}{4}(1-e^{-2x^2})$
  3. $y(x)=\frac{1}{4}(1-e^{2x^2})$
  4. $y(x)=\frac{1}{4}(1-\cos x)$
in Calculus
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Rearranging the terms in the differential equation, we get :

 

$\frac{dy}{1-4y}=xdx$

 

Integrating on both sides, we get:

 

$\int \frac{dy}{1-4y}=\int xdx$

 

$\Rightarrow$   $\frac{1}{-4}\times ln(1-4y)=\frac{x^{2}}{2}+c$

 

Now, using the condition given in the question i.e.  $y(0)=0$,  we get  $c=0$

 

$\therefore$   We get  the equation as    $\frac{1}{-4}\times ln(1-4y)=\frac{x^{2}}{2}$

 

$\Rightarrow$    $ln(1-4y)=-2x^{2}$              $\Rightarrow$       $e^{-2x^{2}}=1-4y$

 

$\Rightarrow$       $y(x)=\frac{1}{4}(1-e^{-2x^{2}})$

 

Option B is the correct answer.

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