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

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

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

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

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

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

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