How do you rewrite the differential equation ( $1+e^x$ )dy = ye^x dx into the standard form of a first-order linear equation?

Q: How do you rewrite the differential equation (1+e^x)dy = ye^x dx into the standard form of a first-order linear equation?

Divide both sides by dx and (1+e^x) to isolate dy/dx, yielding dy/dx = (ye^x)/(1+e^x). Then, move the term containing y to the left side to match the standard linear structure y' + p(x)y = q(x). This results in dy/dx - (e^x/(1+e^x))y = 0.

Divide both sides by dx and (

1+e^x

) to isolate

dy/dx

, yielding

dy/dx

= (ye^x)/(

1+e^x

). Then, move the term containing y to the left side to match the standard linear structure y' +

p(x)y = q(x)

. This results in

dy/dx - (e^x/(1+e^x))y = 0

Conditions

The equation is a first-order differential equation.
( $1+e^x$ ) is never zero, so division is valid.

Reasoning, step by step

Start with the equation: ( $1+e^x$ )dy = ye^x dx.
Divide by $dx(1+e^x)$ : $dy/dx$ = (ye^x)/( $1+e^x$ ).
Rearrange terms to the left side: $dy/dx - (e^x/(1+e^x))y = 0$ .
Identify $p(x) = -e^x/(1+e^x)$ and $q(x) = 0$ .

Example

The video shows this exact transformation, highlighting how to identify the coefficient function $p(x)$ and the nonhomogeneous term $q(x)$ after rearrangement.

Common misconceptions

Forgetting to include the negative sign when identifying $p(x)$ .
Assuming $q(x)$ must be nonzero; in this case, it is zero.

Watch the explanation

BilibiliDifferential equations: how to quickly distinguish Separable equations, Linear equations, and differential equations with variables swapped about X

8:48 – 10:00Watch this moment ↗

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Answers are generated from source material and independently checked. Consult the original video or creator if something is unclear.

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