Find the solution to the initial value problemy′−e−t+5y=−5, y(0)=−1.𝑦′−𝑒−𝑡+5𝑦=−5, 𝑦(0)=−1.
Question
Find the solution to the initial value problemy′−e−t+5y=−5, y(0)=−1.𝑦′−𝑒−𝑡+5𝑦=−5, 𝑦(0)=−1.
Solution
The given differential equation is a first order linear differential equation. The general form of such an equation is y′ + p(t)y = g(t). In this case, p(t) = 5 and g(t) = e^-t - 5.
Step 1: Find the integrating factor e^(∫p(t) dt) = e^(∫5 dt) = e^(5t).
Step 2: Multiply every term of the differential equation by the integrating factor to get e^(5t)y′ - e^(4t) + 5e^(5t)y = -5e^(5t).
Step 3: The left-hand side of this equation is now the derivative of (e^(5t)y) with respect to t. So, we have (e^(5t)y)' = -5e^(5t) + e^(4t).
Step 4: Integrate both sides with respect to t to find y. ∫(e^(5t)y)' dt = ∫(-5e^(5t) + e^(4t)) dt.
Step 5: This gives e^(5t)y = -e^(5t) + 1/5e^(4t) + C, where C is the constant of integration.
Step 6: Finally, solve for y to get y = -1 + 1/5e^(-t) + Ce^(-5t).
Step 7: Use the initial condition y(0) = -1 to find C. -1 = -1 + 1/5 + C, so C = -1/5.
Therefore, the solution to the initial value problem is y = -1 + 1/5e^(-t) - 1/5e^(-5t).
Similar Questions
To find the 𝑦y value of the equation 𝑦=5𝑥y=5x when 𝑥=−1x=−1:Given the equation:𝑦=5𝑥y=5xSubstitute 𝑥=−1x=−1 into the equation:𝑦=5(−1)y=5(−1)𝑦=−5y=−5So, the 𝑦y value when 𝑥=−1x=−1 is −5−5.
Analyse the conditions of Theorem 2.7 for the IVPy′(x) = 5 yx , y(x0) = y0 .For which values of x0, y0 would you expect a unique solution satisfying the BC?Find the general solution to the ODE. Is there a unique solution in a region aboutthe following initial points:(a) y(1) = 1? (b) y(1) = −1? (c) y(−1) = 0? (d) y(0) = y0?For the initial conditions y(1) = −1 show that a unique solution can be defined onx > 0 and exhibit the non-uniqueness of the solution for x ⩾ 0.
Write the point-slope form of the equation of the line that passes through the point (-1, 5) and has a slope of -1. Include your work in your final answer. Type your answer in the box provided to submit your solution.
Is (−5;−154)(−5;−154) a solution to the equation y=−14x−5𝑦=−14𝑥−5?Answer:The correct solution is:
Instructions: Rewrite the equation in Slope-Intercept Form.y−2=−25(x−5)𝑦−2=−25(𝑥−5)
Upgrade your grade with Knowee
Get personalized homework help. Review tough concepts in more detail, or go deeper into your topic by exploring other relevant questions.