1. Solve the differential equation y' = 4 sint/y with initial condition y(0) = 2. 2. Find the...

Question:

1. Solve the differential equation {eq}{y}' = \frac{4\ sin\ t}{y} {/eq} with the initial condition {eq}y(0) = 2 {/eq}.

2. Find the general solution of the differential equation {eq}{y}' = e^{4t} - y {/eq}.

Differential Equation

This question is from the differential equation and we have to find out the differential equation in the first part at the given point and we have to find out the general solution

Answer and Explanation:

{eq}1. \ y'=\frac{4\sin(t)}{y}\\ \Rightarrow \ \frac{dy}{dt}= \frac{4\sin(t)}{y}\\ \Rightarrow \ ydy=4\sin(t)dt\\ \text{Integrating both sides}\\ \Rightarrow \ \int{y}dy=4\int\sin(t)dt\\ \Rightarrow \ \frac{1}{2}y^{2}=-4cos(t)+C\\ \Rightarrow \ y^{2}=-8\cos(t)+2C\\ \Rightarrow \ y=\sqrt{2C-8\cos(t)}\\ since,\\ \Rightarrow \ y(0)=2\\ \Rightarrow \ 2=\sqrt{2C-8}\\ \Rightarrow \ 4=2C-8\\ \Rightarrow \ C=6\\ \Rightarrow \ y=\sqrt{12-8\cos(t)}\\ 2. \ y'=e^{4t}-y\\ \Rightarrow \ \frac{dy}{dt}+y=e^{4t}\\ \text{On comparing with }\frac{dy}{dt}+Py=Q\\ \Rightarrow \ P=1, \ Q=e^{4t}\\ \text{Integrating Factor }=e^{\in{tdt}} \Rightarrow \ e^{t}\\ \text{Hence,}\\ \Rightarrow \ y\cdotp{e^{t}}=\int{e^{4t}}\cdotp{e^{t}}dt\\ \Rightarrow \ y\cdotp{e^{t}}=\frac{e^{5t}}{5}+C\\ \Rightarrow \ y=\frac{e^{4t}}{5}+Ce^{-t}\\ {/eq}


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Evaluating Definite Integrals Using the Fundamental Theorem

from AP Calculus AB: Exam Prep

Chapter 16 / Lesson 2
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