Separable Differential Equations -1 worksheet - Free Printable

Let's solve each of these separable differential equations step by step.

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Problem 1:

Given:
$$ y' = (x^2 + 1)y $$

This is a separable differential equation. We can write it as:
$$
\frac{dy}{dx} = (x^2 + 1)y
$$

Separate variables:
$$
\frac{dy}{y} = (x^2 + 1)\,dx
$$

Now integrate both sides:
$$
\int \frac{1}{y}\,dy = \int (x^2 + 1)\,dx
$$

Left side:
$$
\ln|y| = \int x^2\,dx + \int 1\,dx = \frac{x^3}{3} + x + C
$$

Exponentiate both sides:
$$
|y| = e^{\frac{x^3}{3} + x + C} = e^C \cdot e^{x + \frac{x^3}{3}}
$$

Let $ c = e^C $ (positive constant), so:
$$
y = c e^{x + \frac{1}{3}x^3}
$$

✔ Answer: B
$$
\boxed{y = ce^{x + \frac{1}{3}x^3}}
$$

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Problem 2:

Given:
$$ y' = 2x(y - 1) $$

Rewrite:
$$
\frac{dy}{dx} = 2x(y - 1)
$$

Separate variables:
$$
\frac{dy}{y - 1} = 2x\,dx
$$

Integrate both sides:
$$
\int \frac{1}{y - 1}\,dy = \int 2x\,dx
$$

Left side:
$$
\ln|y - 1| = x^2 + C
$$

Exponentiate:
$$
|y - 1| = e^{x^2 + C} = e^C \cdot e^{x^2}
$$

Let $ c = e^C $, so:
$$
y - 1 = \pm c e^{x^2} \Rightarrow y = 1 + c e^{x^2}
$$
(Note: $ \pm c $ can be absorbed into a new constant $ c $, which can now be any real number.)

✔ Answer: B
$$
\boxed{y = 1 + ce^{x^2}}
$$

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Problem 3:

Given:
$$ y' = 2x^2 y^2 $$

Rewrite:
$$
\frac{dy}{dx} = 2x^2 y^2
$$

Separate variables:
$$
\frac{dy}{y^2} = 2x^2\,dx
$$

Integrate both sides:
$$
\int y^{-2}\,dy = \int 2x^2\,dx
$$

Left side:
$$
\int y^{-2}\,dy = -y^{-1} + C_1
$$

Right side:
$$
\int 2x^2\,dx = 2 \cdot \frac{x^3}{3} = \frac{2}{3}x^3 + C_2
$$

So:
$$
-\frac{1}{y} = \frac{2}{3}x^3 + C
$$

Multiply both sides by -1:
$$
\frac{1}{y} = -\frac{2}{3}x^3 - C
$$

Let $ C' = -C $, then:
$$
\frac{1}{y} = -\frac{2}{3}x^3 + C'
$$

Now solve for $ y $:
$$
y = \frac{1}{C' - \frac{2}{3}x^3}
$$

We can write this as:
$$
y = \frac{1}{c - \frac{2}{3}x^3}, \quad \text{where } c = C'
$$

But none of the options match exactly this form.

Wait — let's check the options:

A- $ y = \dfrac{-1}{\frac{2}{3}x^3 + c} $

B- $ y = \dfrac{1}{3x^3 + c} $

C- $ y = \dfrac{3}{x^3 + c} $

D- $ y = \dfrac{1}{\frac{2}{5}x^3 + c} $

Our solution:
$$
y = \frac{1}{c - \frac{2}{3}x^3} = \frac{1}{-\left(\frac{2}{3}x^3 - c\right)} = \frac{-1}{\frac{2}{3}x^3 - c}
$$

Let’s rename $ -c $ to just $ c $, then:
$$
y = \frac{-1}{\frac{2}{3}x^3 + c}
$$

✔ This matches Option A

So:
$$
\boxed{\text{A}} \quad y = \frac{-1}{\frac{2}{3}x^3 + c}
$$

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✔ Final Answers:

1. B – $ y = ce^{x + \frac{1}{3}x^3} $
2. B – $ y = 1 + ce^{x^2} $
3. A – $ y = \dfrac{-1}{\frac{2}{3}x^3 + c} $

Let me know if you'd like explanations in a different format!