Question

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dy/dx +1 = e^x+y​

Answer 1

★Given:-

• $\displaystyle \sf \frac{dy}{dx} +1 = e^{x+y}$

★Solution:-

Putting x+y = t in the given equation,

x + y = t

Differentiating w.r.t x,

$\displaystyle \implies \sf 1+\frac{dy}{dx} = \frac{dt}{dx}$

According to question,

$\displaystyle \sf \implies \frac{dt}{dx} = e^{x+y}$

[As x+y=t]

$\displaystyle \sf \implies \frac{dt}{dx} = e^t$

$\displaystyle \sf \implies e^{-t} dt = dx$

Integrating on both sides,

$\displaystyle \sf \implies \int dx =\int e^{-t} dt$

$\displaystyle \sf \implies x = -e^{-t}+c$

$\displaystyle \sf \implies x-c = -e^{-t}$

$\displaystyle \sf \implies (x-c)e^t = -1$

Substituting the value of 't',

$\displaystyle \sf \implies (x-c)e^{x+y} = -1$

$\displaystyle \implies \underline{\boxed{\sf (x-c)e^{x+y}+1 = 0}}$

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Answer 2

Given : -

The differential equation is ,

$\frac{dy}{dx} + 1 = e^{x+y}$

Solution : -

Let us take , $( x + y ) = V$

Differentiating the above relation with respect to ' x ' , we get

⇒ $\frac{d( x+y )}{dx} = \frac{dV}{x}$

⇒ $1 + \frac{dy}{dx} = \frac{dV}{dx}$

⇒ $\frac{dy}{dx} = \frac{dV}{dx} -1$   _____(1)

Now ,

$\frac{dy}{dx} + 1 = e^{x + y}$

⇒ $\frac{d V}{dx} - 1 +1 = e^{V}$

⇒ $\frac{dV}{dx} = e^{V}$

Now we can use the variable separable method to solve the differential equation.

⇒ $\frac{dV}{e^{V}} = dx$

Taking integrals on both the sides ,

⇒ $\int {dV * e^{-V} } \ = \int dx$

⇒ $-e^{-V} + c = x$       ( ∵ $\int{e^{-x} } \, dx = -e^{-x}$ )

⇒ $e^{-V} +x = c$

⇒ $e^{-(x+y)} + x = c$       [∵ from equation (1) ]

Therefore the solution the given differential equation is,

$e^{-(x+y)} + x = c$