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**Hint:**By using the given slope , we get a linear differential equation of the form $\dfrac{{dy}}{{dx}} + Py = Q$

To solve such equations , first we need to find the integrating factor $I.F = {e^{\int {Pdx} }}$ and the general solution of the linear differential equation is given by $y{e^{\int {Pdx} }} = \int {Q{e^{\int {Pdx} }} + c} $. As the line passes through the orgin , we can substitute (0,0) in the obtained equation to get the value of c . And this gives the equation of the tangent.

**Complete step by step answer:**

Step 1:

It is given that the slope of the tangent to the curve is y+2x. We know that the slope of the tangent at any point on the curve is $\dfrac{{dy}}{{dx}}$.

So, now from the given condition, we have that

$ \Rightarrow \dfrac{{dy}}{{dx}} = y + 2x$

Now let’s bring the y to the left-hand side

$ \Rightarrow \dfrac{{dy}}{{dx}} - y = 2x$…………………(1)

Step 2:

We know that the general form of the linear differential equation is,

.$\dfrac{{dy}}{{dx}} + Py = Q$.

Now equation (1) is of the form of the linear differential equation where $P = - 1,Q = 2x$

Step 3:

Now we need to solve our differential equation to obtain our required equation.

To solve linear differential equations of this kind we need to find the integrating factor first .

$ \Rightarrow I.F = {e^{\int {Pdx} }}$

Here P=-1

$\begin{gathered}

\Rightarrow I.F = {e^{\int { - 1dx} }} \\

\Rightarrow I.F = {e^{ - \int {dx} }} \\

\\

\end{gathered} $

Now integrating dx in terms of x ,we get

$I.F = {e^{ - x}}$

Step 4:

The general solution is given by

$y{e^{\int {Pdx} }} = \int {Q{e^{\int {Pdx} }} + c} $

Here,we know that ${e^{\int {Pdx} }} = {e^{ - x}}{\text{ and }}Q = 2x$

$ \Rightarrow y{e^{ - x}} = \int {2x.{e^{ - x}}dx + c} $…………….(3)

Step 5:

Now we need to integrate $2\int {x{e^{ - x}}dx} $

Since we have two different expressions $x{\text{ and }}{e^{ - x}}$ as the integrand .we can use integration by parts

$ \Rightarrow \int {udv = uv - \int {vdu} } $

Here our u=x and dv =${e^{ - x}}$

From this our du is obtained by differenting u

$ \Rightarrow du = dx$

And v is obtained by integrating dv

$ \Rightarrow v = \int {dv = \int {{e^{ - x}}dx} = - {e^{ - x}} + c} $

Now lets substitute the values

$ \Rightarrow \int {x{e^{ - x}}dx = x* - {e^{ - x}} - \int { - {e^{ - x}}} } dx$

Now lets integrate $\int {{e^{ - x}}dx} $ again

$ \Rightarrow \int {{e^{ - x}}dx = - {e^{ - x}} + c} $

Therefore

$\begin{gathered}

\Rightarrow \int {x{e^{ - x}}dx = x{e^{ - x}} + ( - {e^{ - x}}) + c} \\

\Rightarrow \int {x{e^{ - x}}dx} = x{e^{ - x}} - {e^{ - x}} + c \\

\end{gathered} $

Step 6

Now lets substitute the above equation in equation (3)

$\begin{gathered}

\Rightarrow y{e^{ - x}} = 2(x{e^{ - x}} - {e^{ - x}}) + c \\

\Rightarrow y{e^{ - x}} = {e^{ - x}}(2x - 2) + c \\

\end{gathered} $

Now lets multiply by ${e^{ - x}}$ on both sides

$ \Rightarrow y = 2x - 2 + c{e^x}$………………(4)

This is the equation of the curve but it is given that the curve passes through the origin

(x,y)=(0,0)

$\begin{gathered}

\Rightarrow 0 = 2(0) - 2 + c{e^0} \\

\Rightarrow 0 = 0 - 2 + c \\

\Rightarrow c = 2 \\

\end{gathered} $

Substituting the value of c in (4)

$\begin{gathered}

\Rightarrow y = 2x - 2 + 2{e^x} \\

\Rightarrow 2x - y - 2 + 2{e^x} = 0 \\

\end{gathered} $

This is the required equation of the curve.

**Additional information**

1) For the circle ${x^2} + {y^2} = {a^2}$, the equation of the tangent whose slope is ‘m’, is given by $y = mx \pm a\sqrt {1 + {m^2}} $

2) This equation is referred to as the ‘slope form’ of the tangent. Given the slope, we can obtain the equation of the tangent

3) The condition for a given line to touch a circle is: Distance of the line from the center of the circle, must be equal to its radius. We’ll refer to this as the ‘condition of tangency’.

**Note:**Many students tend to stop the sum at equation (4) but since we are given that it passes through a point we need to find c using it and substitute it in the equation (4) to find the curve.

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