Differential Equations - dy/dx = f(x)

Differential equations of the form \(\frac{dy}{dx}=f(x)\) are very common and easy to solve. The following shows how to do it:

Step 1
First we multiply both sides by \(dx\) to obtain

\[dy=f(x)~dx.\]

Step 2
Then we take the integral of both sides to obtain

\[\begin{align} \int dy&=\int f(x)~dx\\ y+C'&=\int f(x)~dx\\ \Rightarrow y&=\int f(x)~dx, \end{align}\]

where \(C'\) is the integral constant of \(\int dy.\) We can neglect \(C'\) since another integral constant \(C\) is hidden in the integral \(\int f(x)~dx.\)

Therefore, we can conclude that if \(\frac{dy}{dx}=f(x),\) then \(y=\int f(x)~dx+C.\)

Find \(y\) in terms of \(x\) where \(\frac{dy}{dx}=x.\)

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We have

\[\begin{align} \frac{dy}{dx}&=x\\ dy&=x~dx\\ \int dy&=\int x~dx\\ y&=\frac{1}{2}x^2+C.\ _\square \end{align}\]

Find \(y\) in terms of \(x\) where \(\dfrac{dy}{dx} = \frac {x}{\sqrt{x^2 + 9}}\) and \(y(4)=2.\)

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We have

\[\begin{align} \frac{dy}{dx}&=\frac{x}{\sqrt{x^2+9}}\\ dy&=\frac{x}{\sqrt{x^2+9}}dx\\ \int dy&=\int\frac{x}{\sqrt{x^2+9}}dx\\ y&=\sqrt{x^2+9}+C. \end{align}\]

Substituting \(y(4)=2\) gives

\[2=\sqrt{4^2+9}+C=5+C\Rightarrow C=-3,\]

and therefore our answer is

\[y=\sqrt{x^2+9}-3.\ _\square\]

References

Editorial: PHH Ecuaciones diferenciales elementales y problemas con condiciones en la frontera authors: C.H Edwards, David Penney