The solution of d y d x + 1 = e x + y is Options: A . &nbsp e − ( x + y ) + x + c = 0 B . &nbsp e − ( x + y ) − x + c = 0 C . &nbsp e x + y + x + c = 0 D . &nbsp e x + y − x + c = 0

Answer: Option A : A x + y = z ⇒ 1 + d y d x = d z d x d y d x + 1 = e x + y ⇒ d y d x = e − z d z = d x ⇒ ∫ e − z d z = ∫ d x ⇒ − e − z = x + c ⇒ e − ( x + y ) + x + c = 0.

The solution of dydx+1=ex+y is

Discussion Forum : Differential Equations

Question -

The solution of $\frac{d y}{d x} + 1 = e^{x + y}$ is

Options:

A .

e^{- (x + y)} + x + c = 0

B .

e^{- (x + y)} - x + c = 0

C .

e^{x + y} + x + c = 0

D .

e^{x + y} - x + c = 0

Answer: Option A
:
A

x + y = z \Rightarrow 1 + \frac{d y}{d x} = \frac{d z}{d x} \frac{d y}{d x} + 1 = e^{x + y} \Rightarrow \frac{d y}{d x} = e^{- z} d z = d x \Rightarrow \int e^{- z} d z = \int d x \Rightarrow - e^{- z} = x + c \Rightarrow e^{- (x + y)} + x + c = 0.

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Discussion Forum : Differential Equations

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