Question

The solution of the initial value problem $\frac{dy}{dx} = -6xy \ ; \quad y(0) = 7$ is given by}

• A. $y(x) = 7 e^{-3x^2}$
• B. $y(x) = 3 e^{-7x^2}$
• C. $y(x) = 7 e^{-x^2}$
• D. $y(x) = 3 e^{-x^2}$

Hint:

To verify the solution quickly, test the initial condition $y(0) = 7$ directly in the options:
- For Option (A): $y(0) = 7 e^0 = 7$ (valid).
- For Option (C): $y(0) = 7 e^0 = 7$ (valid).
Then differentiate Option (A) to see if it matches: $y' = 7 e^{-3x^2}(-6x) = -6xy$ (correct).

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
A first-order ordinary differential equation can be solved by grouping all terms of $y$ on one side and all terms of $x$ on the other side through separation of variables.

Step 2: Detailed Explanation:
We are given the differential equation:
\[ \frac{dy}{dx} = -6xy \]
Separate the variables by dividing both sides by $y$ and multiplying by $dx$:
\[ \frac{1}{y} \, dy = -6x \, dx \]
Integrate both sides of the equation:
\[ \int \frac{1}{y} \, dy = \int -6x \, dx \]
\[ \ln|y| = -3x^2 + C_1 \]
Exponentiate both sides to solve for $y$:
\[ y(x) = e^{-3x^2 + C_1} = e^{C_1} \cdot e^{-3x^2} \]
Let $C = e^{C_1}$ be an arbitrary integration constant:
\[ y(x) = C e^{-3x^2} \]
Now, apply the initial condition $y(0) = 7$:
\[ 7 = C e^{-3(0)^2} \]
\[ 7 = C \cdot e^0 \implies C = 7 \]
Substitute $C = 7$ back into the general solution:
\[ y(x) = 7 e^{-3x^2} \]
Therefore, the solution to the initial value problem is $y(x) = 7 e^{-3x^2}$.

Step 3: Final Answer
The correct option is (A).