"Find the differential equations of the following. Show..." was answered by our experts

Gregory Emery

Answered question

2021-12-31

Find the differential equations of the following. Show complete solution.

\frac{d y}{d x} + y = e^{2 x}

Answer & Explanation

Janet Young

Beginner2022-01-01Added 32 answers

Step 1
Given:

\frac{d y}{d x} + y = e^{2 x}

To Find: To find the general solution of the given differential equation.
Step 2
Solution:

\frac{d y}{d x} + y = e^{2 x}

\Rightarrow \frac{d y}{d x} + 1 y = e^{2 x}

(1)
We observe that equation (1) is a first order linear ordinary differential

= q

equation of the form,

\frac{d y}{d x} + p y

Where,

p = 1 and q = e^{2 x}

\Rightarrow Integrating factor = e^{\int p d x} = e^{\int 1 d x} = e^{x}

Multiplying equation (1) by the Integrating factor

= e^{x}

, we get

e^{x} \frac{d y}{d x} + e^{x} y = e^{x} (e^{2 x})

e^{x} \frac{d y}{d x} + e^{x} y = e^{x + 2 x}

\Rightarrow e^{x} \frac{d y}{d x} + e^{x} y = e^{3 x}

\Rightarrow \frac{d}{d x} (y . e^{x}) = e^{3 x}

\Rightarrow d (y . e^{x}) = e^{3 x} d x

Integrating both the sides we get

\int d (y . e^{x}) = \int e^{3 x} d x

\Rightarrow y . e^{x} = (\frac{e^{3 x}}{3}) + c

, where

c =

constant of integration

\Rightarrow y . e^{x} = \frac{1}{3} e^{3 x} + c

\Rightarrow y = \frac{1}{e^{x}} (\frac{1}{3} e^{3 x} + c)

\Rightarrow y = \frac{1}{3} (\frac{e^{3 x}}{e^{x}}) + \frac{c}{e^{x}}

Hector Roberts

Beginner2022-01-02Added 31 answers

Given

\frac{d y}{d x} + y = e^{2 x}

It is in the linear differential equation form of

\frac{d y}{d x} + P y = Q

Here,

P = 1 and Q = e^{2 x}

We first find the integrating factor(IF)

I = e^{\int p d x} = e^{\int 1 . d x} = e^{x}

Then we multiply the differential equation with IF to get

y \times I F = \int Q \times I F . d x + C

\Rightarrow e^{x} y = e^{x} (e^{2 x}) + C

\Rightarrow e^{x} (y) = \frac{e^{3 x}}{3} + C

\Rightarrow y = \frac{e^{3 x}}{3} e^{x} + c . e^{- x}

General solution,

y = \frac{e^{2 x}}{3} + c . e^{- x}

user_27qwe

Skilled2022-01-05Added 375 answers

$\begin{matrix} d y / d x + y = e^{2 x} \\ d y / d x + P y = Q \\ P = 1 a n d Q = e^{2 x} \\ I = e^{\int p d x} = e^{\int 1. d x} = e^{x} \\ y \times I F = \int Q \times I F . d x + C \\ \Rightarrow e^{x} y = e^{x} (e^{2 x}) + C \\ \Rightarrow e^{x} (y) = e^{3 x} / 3 + C \\ \Rightarrow y = e^{3 x} / 3 e^{x} + c . e^{- x} \\ General solution, y = e^{2 x} / 3 + c . e^{- x} \end{matrix}$

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