Question

Solve the initial-value problem dy/dx = e²x - 3y, y(0) = 1
y(x) = ?

Answer 1

To solve the initial-value problem dy/dx = e²x - 3y, y(0) = 1, you need to separate the variables by bringing all terms with y to one side and all terms with x to the other side. Integrate both sides and simplify the integrals. Apply the initial condition to find the value of C. Exponentiate both sides to eliminate the natural logarithm and solve for y by considering both positive and negative solutions.

Step-by-step explanation:

To solve the initial-value problem dy/dx = e²x - 3y, y(0) = 1, we need to find the function y(x). Here are the step-by-step instructions:

1. Separate the variables by bringing all terms with y to one side and all terms with x to the other side: dy - 3y dy = e²x dx
2. Integrate both sides: ∫(1/y - 3) dy = ∫e²x dx
3. Simplify the integrals: ln|y - 3y²/2| = e²x + C
4. Apply the initial condition y(0) = 1 to find C: ln|1 - 3/2| = 0 + C
5. Determine the value of C: ln(1/2) = C
6. Write the final equation: ln|y - 3y²/2| = e²x + ln(1/2)
7. Exponentiate both sides to eliminate the natural logarithm: |y - 3y²/2| = e²x * (1/2)
8. Solve for y by considering both positive and negative solutions: y - 3y²/2 = ±(e²x / 2)
9. Simplify the equation: y - 3y²/2 = ±e²x / 2
10. Now, you can solve for y using algebraic techniques or numerical methods to obtain the final function y(x).