Exercises - Separable Differential Equations

  1. Solve $\displaystyle{x^2+4-y^3 \frac{dy}{dx} = 0}$

    First we move the term involving $y$ to the right side to begin to separate the $x$ and $y$ variables.

    $$x^2 + 4 = y^3 \frac{dy}{dx}$$

    Then, we multiply both sides by the differential $dx$ to complete the separation.

    $$(x^2+4) \, dx = y^3 \, dy$$

    Taking the integral of both sides, we have

    $$\int (x^2 + 4) \, dx = \int y^3 \, dy$$

    Doing the integration and remembering that the resulting constants can be combined to a single arbitrary $C$ gives us an implicit definition of $y$.

    $$\frac{1}{3}x^3 + 4x = \frac{1}{4}y^4 + C$$

    We might clean our solution up a bit by isolating the constant on the right side.

    $$4x^3 + 48x - 3y^4 = C$$
  2. Solve $\displaystyle{\frac{dy}{dx} = \frac{5x^2y^2+y^2}{x^2y^5+4x^2}}$ where $y(1)=2$

    Factoring the expression on the left tells us

    $$\frac{dy}{dx} = \frac{y^2 (5x^2 + 1)}{x^2 (y^5 + 4)}$$

    These factors can then be separated into those involving $x$ and those involving $y$

    $$\frac{dy}{dx} = \left(\frac{y^2}{y^5+4}\right)\left(\frac{5x^2+1}{x^2}\right)$$

    We complete the separation by moving the expressions in $x$ (including $dx$) to one side of the equation, and the expressions in $y$ (including $dy$) to the other.

    $$\frac{y^5+4}{y^2} dy = \frac{5x^2 + 1}{x^2} dx$$

    After rewriting both sides in a more "integration-friendly" form (setting things up for the power-rule), we then take the integral of both sides.

    $$\int (y^3 + 4y^{-2}) \, dy = \int (5 + x^{-2}) \, dx$$

    Doing the integration, and combining the appropriate constants into a single arbitrary $C$ gives us an implicit definition of $y$

    $$\frac{1}{4}y^4 - 4y^{-1} = 5x - x^{-1} + C$$

    We clean this general solution up by writing it with only positive exponents and isolating $C$ on one side.

    $$\frac{1}{4}y^4 - \frac{4}{y} - 5x + \frac{1}{x} = C$$

    To find the particular solution where $y(1)=2$, we simply substitute $x=1$ and $y=2$ into this general solution to find $C$.

    $$\frac{1}{4} \cdot 2^4 - \frac{4}{2} - 5 \cdot 1 + \frac{1}{1} = C$$

    Solving the above, we find $C = -2$. Thus, our particular solution is given by

    $$\frac{1}{4}y^4 - \frac{4}{y} - 5x + \frac{1}{x} = -2$$
  3. Solve $\displaystyle{\frac{dy}{dx} = \frac{\csc^2 (2x) \cot (2x)}{4y}}$

    Note that everything is already factored on the left and right sides of the equation, so we can immediately move things around so that the expressions involving $y$ (including $dy$) are on one side of the equation and the expressions involving $x$ (including $dx$) are on the other.

    $$4y \, dy = \csc^2 (2x) \cot (2x) \, dx$$

    Now we integrate both sides...

    $$\int 4y \, dy = \int \csc^2 (2x) \cot (2x) \, dx$$

    Note that we'll need to employ $u$-substitution to integrate the right side. A good choice for the substitution is $u= \cot(2x)$, which leads to $du = -2\csc^2(2x) \, dx$, and consequently $(-1/2) \, du = \csc^2 (2x) \, dx$. This means the integral on the right can be rewritten as

    $$-\frac{1}{2}\int u \, du = -\frac{1}{4}u^2 + C = -\frac{1}{4} \cot^2 (2x) + C$$

    This tells us that

    $$2y^2 = -\frac{1}{4}\cot^2 (2x) + C$$

    Cleaning things up by isolating the $C$ on one side, we have the solution

    $$2y^2 + \frac{1}{4}\cot^2 (2x) = C$$
  4. Solve $\displaystyle{\frac{\sqrt{x-4}}{y^2} = \frac{dy}{dx}}$ where $y(4) = -1$

    $2(x-4)^{3/2} - y^3 = 1$

  5. Solve $\displaystyle{\frac{x^2y(x^2+4)^2}{xy^2-xy^3} = \frac{dy}{dx}}$

    $(x^2 + 4)^3 - 3y^2 + 2y^3 = C$

  6. Solve $\displaystyle{\frac{x+4}{\sqrt{xy}} = \frac{dy}{dx}}$

    $\displaystyle{\frac{2}{3} x^{3/2} + 8x^{1/2} + \frac{2}{3} y^{3/2} = C}$

  7. Solve $\displaystyle{x^2-5x+4 = \frac{dy}{dx}}$ where $y(0) = -2$

    $\displaystyle{\frac{1}{3} x^3 - \frac{5}{2} x^2 + 4x - y = 2}$

  8. Solve $\displaystyle{\frac{dy}{dx} = \frac{\sqrt{4-y} \cos \sqrt{x}}{y^2\sqrt{x}}}$

  9. Solve $\displaystyle{\frac{dy}{dx} = \frac{x+4}{x^3 \csc (2y) \cot (2y)}}$