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For the differential equation dy/dt + 5y = 0, with y(0) = 1, the general solution is
- a)5e5t
- b)e-5t
- c)e5t
- d)up>(d) 5e-5t
Correct answer is option 'B'. Can you explain this answer?
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For the differential equation dy/dt + 5y = 0, with y(0) = 1, the gen...
Given Differential Equation:
dy/dt + 5y = 0
Initial Condition:
y(0) = 1
General Solution:
The general solution of the given differential equation can be found by solving it using the method of separation of variables.
Step 1: Separate the variables.
dy/y = -5 dt
Step 2: Integrate both sides.
∫(dy/y) = ∫(-5 dt)
Step 3: Evaluate the integrals.
ln|y| = -5t + C1
Step 4: Solve for y.
Taking the exponential of both sides, we get:
|y| = e^(-5t + C1)
Using the absolute value, we can write it as:
y = ±e^(-5t + C1)
Step 5: Simplify the constant.
Let C = ±e^C1, where C is a non-zero constant.
The general solution can now be written as:
y = Ce^(-5t)
Applying Initial Condition:
Using the initial condition y(0) = 1, we can find the specific value of the constant C.
y(0) = Ce^(-5*0)
1 = Ce^0
1 = C
Therefore, the specific solution is:
y = e^(-5t)
Correct Answer:
Option B) e^(-5t)
dy/dt + 5y = 0
Initial Condition:
y(0) = 1
General Solution:
The general solution of the given differential equation can be found by solving it using the method of separation of variables.
Step 1: Separate the variables.
dy/y = -5 dt
Step 2: Integrate both sides.
∫(dy/y) = ∫(-5 dt)
Step 3: Evaluate the integrals.
ln|y| = -5t + C1
Step 4: Solve for y.
Taking the exponential of both sides, we get:
|y| = e^(-5t + C1)
Using the absolute value, we can write it as:
y = ±e^(-5t + C1)
Step 5: Simplify the constant.
Let C = ±e^C1, where C is a non-zero constant.
The general solution can now be written as:
y = Ce^(-5t)
Applying Initial Condition:
Using the initial condition y(0) = 1, we can find the specific value of the constant C.
y(0) = Ce^(-5*0)
1 = Ce^0
1 = C
Therefore, the specific solution is:
y = e^(-5t)
Correct Answer:
Option B) e^(-5t)
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For the differential equation dy/dt + 5y = 0, with y(0) = 1, the general solution isa)5e5tb)e-5tc)e5td)up>(d) 5e-5tCorrect answer is option 'B'. Can you explain this answer?
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