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Math  /  Calculus

QuestionLet AA and kk be positive constants. Which of the given functions is a solution to dydt=−k(y+A)\frac{d y}{d t}=-k(y+A) ?
Answer: □\square
Select an option Incorrect y=A+Cekty=−A+Ce−kty=A+C e^{k t} \quad y=-A+C e^{-k t} y=−A+Cekty=-A+C e^{k t} y=−A+Cekty=A−1+Ce−Akty=A+Ce−kty=A−1+CeAkt\begin{array}{l} y=-A+C e^{k t} y=A^{-1}+C e^{-A k t} y=A+C e^{-k t} \\ y=A^{-1}+C e^{A k t} \end{array}

Studdy Solution

STEP 1

1. We are given a first-order linear differential equation: dydt=−k(y+A)\frac{d y}{d t} = -k(y + A).
2. We need to find a function y(t) y(t) that satisfies this differential equation.
3. A A and k k are positive constants, and C C is an arbitrary constant.

STEP 2

1. Rearrange the differential equation.
2. Solve the differential equation using separation of variables.
3. Identify the correct solution from the given options.

STEP 3

Rearrange the given differential equation dydt=−k(y+A)\frac{d y}{d t} = -k(y + A) to separate variables:
dydt=−k(y+A) \frac{d y}{d t} = -k(y + A)
This can be rewritten as:
dydt=−ky−kA \frac{d y}{d t} = -ky - kA

STEP 4

Separate variables by moving all terms involving y y to one side and t t to the other side:
dyy+A=−k dt \frac{d y}{y + A} = -k \, dt
Integrate both sides:
∫1y+A dy=∫−k dt \int \frac{1}{y + A} \, dy = \int -k \, dt
The left side integrates to:
ln⁡∣y+A∣ \ln|y + A|
The right side integrates to:
−kt+C1 -kt + C_1
Thus, we have:
ln⁡∣y+A∣=−kt+C1 \ln|y + A| = -kt + C_1

STEP 5

Solve for y y by exponentiating both sides to remove the natural logarithm:
∣y+A∣=e−kt+C1 |y + A| = e^{-kt + C_1}
This can be rewritten as:
y+A=±eC1e−kt y + A = \pm e^{C_1} e^{-kt}
Let C=±eC1 C = \pm e^{C_1} , which is an arbitrary constant. Therefore, we have:
y+A=Ce−kt y + A = C e^{-kt}
Solve for y y :
y=−A+Ce−kt y = -A + C e^{-kt}

STEP 6

Identify the correct solution from the given options. The derived solution is:
y=−A+Ce−kt y = -A + C e^{-kt}
The correct option is:
y=−A+Ce−kt y = -A + C e^{-kt}
The solution to the differential equation is:
y=−A+Ce−kt \boxed{y = -A + C e^{-kt}}

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