Given the differential equation y'=f(x,y) , suppose it…

Given the differential equation \(y'=f(x,y)\), suppose it admits an integrating factor that depends only on \(x\). Show that the equation is then linear.

1. Set up \(M\,dx+N\,dy=0\)

\(y'=f(x,y)\) rewrites as \(dy-f(x,y)\,dx=0\), i.e. \(M\,dx+N\,dy=0\) with \(M(x,y)=-f(x,y)\), \(N(x,y)=1\).

2. Derive the standard condition for \(\mu(x)\) to exist — proved, not just cited

Suppose \(\mu=\mu(x)\) makes \(\mu M\,dx+\mu N\,dy=0\) exact. Exactness requires \(\partial_y(\mu M)=\partial_x(\mu N)\), i.e. (since \(\mu\) depends on \(x\) alone) \[ \mu\,M_y = \mu' N + \mu\,N_x \quad\Longleftrightarrow\quad \frac{\mu'}{\mu} = \frac{M_y-N_x}{N}.\qquad(\dagger) \] The left side of \((\dagger)\) depends on \(x\) only (because \(\mu=\mu(x)\)). Therefore, for \((\dagger)\) to hold identically over the domain, the right side must also be independent of \(y\) — this is the necessary direction: if \(\mu(x)\) exists, then \((M_y-N_x)/N\) is a function of \(x\) alone. Conversely, if \((M_y-N_x)/N=:g(x)\) is already known to depend on \(x\) only, then \(\mu(x)=\exp\!\big(\int g(x)\,dx\big)\) satisfies \((\dagger)\) by construction, so exactness holds — this is the sufficient direction. Both directions hold, so the textbook statement really is a genuine if-and-only-if, not merely an assumed equivalence.

3. Apply the necessary direction to \(M=-f,\ N=1\)

Here \(N=1\) so \(N_x=0\) and \(M_y=-f_y\), giving \((M_y-N_x)/N=-f_y\). Since only the hypothesis "\(\mu(x)\) exists" is given (not the condition on \((M_y-N_x)/N\)), it is exactly the necessary direction proved in Step 2 that licenses concluding \[ -f_y(x,y) = g(x) \quad\text{for some function \(g\) of \(x\) alone,} \] i.e. \(f_y(x,y)=h(x)\) for \(h=-g\), a function of \(x\) alone. This is the crux inference and it is logically justified (not circular), because it uses only the "only if" half of the genuine biconditional established in Step 2 — precisely the direction the problem's hypothesis supplies.

4. Integrate \(f_y=h(x)\) with respect to \(y\)

Since \(f_y\) does not depend on \(y\), integrating in \(y\) at fixed \(x\) gives \[ f(x,y) = h(x)\,y + \phi(x) \] for an arbitrary function \(\phi(x)\) of \(x\) alone (the "constant" of \(y\)-integration). Writing \(P(x)=-h(x)\), \(Q(x)=\phi(x)\): \[ f(x,y) = -P(x)y + Q(x). \]

5. Conclude linearity

Substituting back into \(y'=f(x,y)\): \[ y' = -P(x)y+Q(x) \quad\Longleftrightarrow\quad y'+P(x)y=Q(x), \] which is precisely the standard form of a first-order linear ODE. (As a check on the converse, note that any equation already of this form indeed has the classical integrating factor \(\mu(x)=e^{\int P(x)dx}\), consistent with the sufficiency direction proved in Step 2 — though the problem itself only asks for the forward direction.)