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Math  /  Calculus

Question(7) USE SEPERRAION OF JARIABLES TO SOLVE dydt=23ty+yy(1)=−5\frac{d y}{d t}=\frac{2}{3 t y+y} \quad y(1)=-5
AT THE FIRST STEP, TIEE SOLUTION IS GIVEN BY f(y)=g(t)+Cf(y)=g(t)+C WHERE f(y)f(y) anD g(t)g(t) DO NOT CONTANN ANY CONSTANS TERCMS AND F HAS a LEAding COEFFICLENT 1/21 / 2. ENTER TIIESE FUNCTIOWS, THE CONSTANT C FOR WIHCLH THE INITIAL CONOTION IS SATISTIED AND THES SLLUTION. f(y)=g(t)=C=y(t)=\begin{array}{l} f(y)= \\ g(t)= \\ C= \\ y(t)= \end{array}

Studdy Solution

STEP 1

1. The differential equation given is separable, allowing us to separate the variables yy and tt.
2. The differential equation is dydt=23ty+y\frac{d y}{d t} = \frac{2}{3 t y + y}.
3. The initial condition is y(1)=−5y(1) = -5.
4. The solution can be written as f(y)=g(t)+Cf(y) = g(t) + C where f(y)f(y) and g(t)g(t) do not contain any constant terms and ff has a leading coefficient 12\frac{1}{2}.

STEP 2

1. Separate the variables yy and tt in the differential equation.
2. Integrate both sides with respect to their respective variables.
3. Determine the constant CC using the initial condition y(1)=−5y(1) = -5.
4. Solve for y(t)y(t) explicitly if possible.

STEP 3

Separate the variables in the differential equation. dydt=23ty+y  ⟹  dy2y=dt3t+1 \frac{d y}{d t} = \frac{2}{3 t y + y} \implies \frac{d y}{2y} = \frac{d t}{3 t + 1}

STEP 4

Integrate both sides of the separated equation. ∫12y dy=∫13t+1 dt \int \frac{1}{2y} \, dy = \int \frac{1}{3t + 1} \, dt

STEP 5

Perform the integration on both sides. ∫12y dy=12∫1y dy=12ln⁡∣y∣ \int \frac{1}{2y} \, dy = \frac{1}{2} \int \frac{1}{y} \, dy = \frac{1}{2} \ln |y| ∫13t+1 dt=13ln⁡∣3t+1∣ \int \frac{1}{3t + 1} \, dt = \frac{1}{3} \ln |3t + 1|

STEP 6

Combine the results of the integrals and include the constant of integration CC. 12ln⁡∣y∣=13ln⁡∣3t+1∣+C \frac{1}{2} \ln |y| = \frac{1}{3} \ln |3t + 1| + C

STEP 7

Solve for the constant CC using the initial condition y(1)=−5y(1) = -5. 12ln⁡∣−5∣=13ln⁡∣3⋅1+1∣+C \frac{1}{2} \ln |-5| = \frac{1}{3} \ln |3 \cdot 1 + 1| + C 12ln⁡5=13ln⁡4+C \frac{1}{2} \ln 5 = \frac{1}{3} \ln 4 + C C=12ln⁡5−13ln⁡4 C = \frac{1}{2} \ln 5 - \frac{1}{3} \ln 4

STEP 8

Combine the results to express y(t)y(t) explicitly if possible. Exponentiate both sides to eliminate the logarithms. 12ln⁡∣y∣=13ln⁡∣3t+1∣+(12ln⁡5−13ln⁡4) \frac{1}{2} \ln |y| = \frac{1}{3} \ln |3t + 1| + \left(\frac{1}{2} \ln 5 - \frac{1}{3} \ln 4\right) ln⁡∣y∣=23ln⁡∣3t+1∣+ln⁡5−23ln⁡4 \ln |y| = \frac{2}{3} \ln |3t + 1| + \ln 5 - \frac{2}{3} \ln 4 ln⁡∣y∣=ln⁡((3t+1)2/3⋅542/3) \ln |y| = \ln \left( (3t + 1)^{2/3} \cdot \frac{5}{4^{2/3}} \right) ∣y∣=(3t+1)2/3⋅542/3 |y| = (3t + 1)^{2/3} \cdot \frac{5}{4^{2/3}} Since yy is negative initially, y=−(3t+1)2/3⋅542/3 y = - (3t + 1)^{2/3} \cdot \frac{5}{4^{2/3}}
The solution is: f(y)=12ln⁡∣y∣g(t)=13ln⁡∣3t+1∣C=12ln⁡5−13ln⁡4y(t)=−(3t+1)2/3⋅542/3\begin{array}{l} f(y)= \frac{1}{2} \ln |y| \\ g(t)= \frac{1}{3} \ln |3 t + 1| \\ C= \frac{1}{2} \ln 5 - \frac{1}{3} \ln 4 \\ y(t)= - (3t + 1)^{2/3} \cdot \frac{5}{4^{2/3}} \end{array}

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