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Solve the differential equation \( \frac{d y}{d x}=x^{3} y \) \( \mathrm{~A} \quad y=A e^{4 x} \) \( \mathrm{~B} \quad y=A e^{x} \) \( \mathrm{C} \quad y=A e^{\frac{x}{4}} \) \( \mathrm{D} \quad y=A e^{\frac{x^{4}}{4}} \)

Ask by Griffiths Lowe. in Malaysia
Mar 08,2025

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Tutor-Verified Answer

Answer

The solution is \( y = A e^{\frac{x^{4}}{4}} \), choice D.

Solution

We start with the differential equation \[ \frac{dy}{dx} = x^3 y. \] **Step 1: Separate the variables.** Divide both sides by \( y \) (assuming \( y \neq 0 \)) and multiply by \( dx \) to obtain \[ \frac{1}{y} \, dy = x^3 \, dx. \] **Step 2: Integrate both sides.** Integrate the left side with respect to \( y \) and the right side with respect to \( x \): \[ \int \frac{1}{y} \, dy = \int x^3 \, dx. \] The integrals are: \[ \ln |y| = \frac{x^4}{4} + C, \] where \( C \) is the constant of integration. **Step 3: Solve for \( y \).** Exponentiate both sides to isolate \( y \): \[ |y| = e^{\frac{x^4}{4} + C} = e^{\frac{x^4}{4}} \cdot e^{C}. \] Let \( A = \pm e^{C} \) (which is an arbitrary constant), so we have: \[ y = A e^{\frac{x^4}{4}}. \] **Conclusion:** Thus, the solution is \[ y = A e^{\frac{x^4}{4}}, \] which corresponds to option D.

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The Deep Dive

To solve the differential equation \( \frac{dy}{dx} = x^3 y \), we can separate the variables. We rewrite it as: \[ \frac{dy}{y} = x^3 dx \] Next, we integrate both sides: \[ \int \frac{dy}{y} = \int x^3 dx \] This gives us: \[ \ln |y| = \frac{x^4}{4} + C \] Exponentiating both sides, we have: \[ |y| = e^C e^{\frac{x^4}{4}} = A e^{\frac{x^4}{4}} \] where \( A = e^C \). Thus, the solution to the differential equation is: \[ y = A e^{\frac{x^4}{4}} \] Looking at the options provided, the correct answer is: \( \mathrm{D} \quad y=A e^{\frac{x^{4}}{4}} \) Now, let’s add some fun facts! The 18th-century mathematician and philosopher, Gottfried Wilhelm Leibniz, co-developed calculus independently from Isaac Newton around the same time, and his notation is still widely used today! Fun fact: Leibniz’s notation for the derivative, \( dy/dx \), is a fundamental part of how we express rates of change! Additionally, this particular type of differential equation is often used in mathematical biology to model populations where the growth rate is proportional to the current size. It’s fascinating how mathematical tools can help us understand and predict real-world phenomena, like how populations grow exponentially in ideal conditions!

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