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Question

The solution of

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

with

y(0) = 0

is:

(A)

y = -\log(2 - e^x)

(B)

y = \log(2 - e^x)

(C)

y = -\log(1 + e^x)

(D)

y = \log(1 - e^x)

Solution Path

Separate

e^{-y} dy = e^x dx

, integrate, apply IC

y(0)=0

to get

y = -\log(2-e^x)

01Question Setup

1/4

Solve

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

with

y(0) = 0

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

02Separate Variables

2/4

Rewrite as

e^{-y} dy = e^x dx

. Both sides are now single-variable.

e^{-y} dy = e^x dx

03Integrate Both SidesKEY INSIGHT

3/4

-e^{-y} = e^x + C

. Using

y(0) = 0

-1 = 1 + C

, so

C = -2

. Therefore

e^{-y} = 2 - e^x

e^{-y} = 2 - e^x

04Final Answer

4/4

y = -\log(2 - e^x)

\boxed{y = -\log(2 - e^x)}

Concepts from this question1 concepts unlocked

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EASY

Rearrange the equation so that all y terms are on one side and all x terms on the other, then integrate both sides

\int \frac{dy}{g(y)} = \int f(x)\,dx + C

The simplest and most direct solving method; recognising when an equation is separable saves significant time

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