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Differential Equations

Section: Mathematics

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Question : 4 of 48

Marks: +1, -0

If the slope of the tangent drawn at any point (

x, y

) on a curve is (

x+y

), then the equation of that curve is

[AP EAPCET 24 May 2025 S1]

$\;y=c e^{x}+1+x$
$\;y=c e^{x}-x$
$\;y=c e^{-x}-1-x$
$\;y=c e^{x}-1-x$

Solution:

Since, the slope of tangent to a curve at any point

(x, y)

is

x+y

.

\therefore \;\; \;\frac{dy}{dx}=x+y \Rightarrow \;\frac{dy}{dx}-y=x

This is a linear differentiation equation

So,

P=-1

and

Q=x

Now,

\text{IF}=e^{\int P\,dx}=e^{\int -d\,x}=e^{-x}

\Rightarrow \;\; e^{-x} \cdot \;\frac{dy}{dx}-e^{-x} y=x e^{-x}

\Rightarrow \;\; \;\frac{dy}{dx} (y \cdot e^{-x}) =x e^{-x}

Integrating both sides, we get

y \cdot e^{-x}\;=\int x e^{-x}\,dx

\;=-x e^{-x}-\int -e^{-x}\,dx

\;=-x e^{-x}+ (-e^{-x}) +c

where

C=

constant of integration

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

\Rightarrow y=-x-1+c e^{x}

, where

c

is arbitrary constant.

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