# The slope of the tangent to the curve at any point is equal to y+2x.If the curve passes through the origin, then find its equation

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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.

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