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Find the particular solution of dy/dx = 2y with the initial condition y(0) = 1.
Find the particular solution of dy/dx = 2y with the initial condition y(0) = 1.
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Practice Questions
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Q1
Find the particular solution of dy/dx = 2y with the initial condition y(0) = 1.
y = e^(2x)
y = e^(2x) + 1
y = 1 + e^(2x)
y = e^(2x) - 1
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The general solution is y = Ce^(2x). Using the initial condition y(0) = 1 gives C = 1, so y = e^(2x).
Questions & Step-by-step Solutions
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Q
Q: Find the particular solution of dy/dx = 2y with the initial condition y(0) = 1.
Solution:
The general solution is y = Ce^(2x). Using the initial condition y(0) = 1 gives C = 1, so y = e^(2x).
Steps: 12
Show Steps
Step 1: Start with the differential equation dy/dx = 2y.
Step 2: Recognize that this is a separable equation, meaning we can separate y and x.
Step 3: Rewrite the equation as dy/y = 2 dx.
Step 4: Integrate both sides. The left side becomes ln|y| and the right side becomes 2x + C (where C is a constant).
Step 5: After integrating, we have ln|y| = 2x + C.
Step 6: To solve for y, exponentiate both sides to get y = e^(2x + C).
Step 7: Rewrite e^(2x + C) as y = e^(2x) * e^C. Let e^C be a new constant, which we can call C (still a constant). So, y = Ce^(2x).
Step 8: Now we have the general solution: y = Ce^(2x).
Step 9: Use the initial condition y(0) = 1 to find the value of C.
Step 10: Substitute x = 0 into the general solution: y(0) = Ce^(2*0) = C.
Step 11: Set C equal to 1 because y(0) = 1. So, C = 1.
Step 12: Substitute C back into the general solution: y = 1 * e^(2x), which simplifies to y = e^(2x).
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