Community Answer

Honor CodeSolved 1 Answer

See More Answers for FREE

Enhance your learning with StudyX

Receive support from our dedicated community users and experts

See up to 20 answers per week for free

Experience reliable customer service

Get Started

Ques: Gnsider the following Initial Value problem\frac{d y}{d x}=5 y+x \quad, y(0)=1Whe Euler's methoo to costimate the selution y(x) on the inferval [0,1] werth 5 stop.8 f^{n}=-\quad Given \quad \frac{d y}{d x}=5 y+x\begin{array}{l}y(0)=1 \\x=0 \Rightarrow y=1\end{array}\frac{d y}{d x}-5 y=xwe knoro that\frac{d y}{d x}+\mu y=\thetawher p, and \theta function of x and antontp=-5, \theta=xNow Intecgeting facter (I.F)\begin{aligned}I \cdot F & =e^{\int p d x} \\& =e^{\int-5 d x}=e^{-5 \int d x}=e^{-5 x} \\I \cdot F & =e^{-5 x}\end{aligned}Now Gereval sel"y(I, F)=\int \theta(I \cdot F) d xput vake of I.F and \thetaSoy\left(e^{-5 x}\right) \doteq \int \frac{x \cdot e^{-5 x}}{I} \cdot d xwe know thatAlgebra furction Nuenter I and expenitial function Nambor III\int I \cdot I d x=I \int \pi d x-\int\left(\frac{d}{d x}(I) \int d x\right) d xNow y\left(e^{-5 x}\right)=x \int e^{-5 x} d x-\left(\frac{d x}{d x}(x) \int e^{-5 x} d x\right) d xWe know that\begin{array}{l}\int e^{a x} d x=\frac{1}{a} e^{a x}+c \quad, \frac{d}{d x}(x)=1 \\\Rightarrow y\left(e^{-5 x}\right)=\frac{x e^{-5 x}}{-5}-\int \cdot \frac{e^{-5 x}}{-5} d x+c \\\Rightarrow y\left(e^{-5 x}\right)=-\frac{1 x}{5} e^{-5 x}+\frac{1}{5}\left(\frac{e^{-5 x}}{5}\right)+c \\\Rightarrow y\left(e^{-5 x}\right)=-\frac{1 x}{5} e^{-5 x}+\left(\frac{-1}{25}\right) e^{-5 x}+c \\\Rightarrow y e^{-5 x}=e^{-5 x}\left(-\frac{1 x}{5}-\frac{1}{25}\right)+c \\\Rightarrow \quad y=\frac{-x}{5}+\frac{1}{25}+c \\\end{array}Now given y(0)=1x=0 \Rightarrow y=1put values in ogn (2)\begin{array}{l}1=\frac{-0}{5}+\left(\frac{-1}{25}\right)+c \\c=1+\frac{1}{25}=\frac{26}{25} \\c=\frac{26}{25}\end{array}Hence\begin{array}{l}y=-\frac{x}{5}-\frac{1}{25}+\frac{26}{25} \\y=-\frac{x}{5}+\frac{25}{25} \\y=-\frac{x}{5}+1 \\y(x)=-\frac{x}{5}+1\end{array} ...