Differential Equations Exercise 4.1 Text Book Solutions

Dear Class 12 Samacheer Kalvi students, here are the Business Maths Chapter 4 Differential Equations Exercise 4.2 Text Book Solutions for your reference and study.

You can find links to the other exercises below.

Important Formulas and Notes in Differential Equations

Text Book Solutions for Differential Equations Exercise 4.1

Text Book Solutions for Differential Equations Exercise 4.3

Text Book Solutions for Differential Equations Exercise 4.4

Text Book Solutions for Differential Equations Exercise 4.5

Text Book Solutions for Differential Equations Exercise 4.6

Exercise 4.2

$1. (i)\frac{dy}{dx}=ae^{y} \\$

$\frac{dy}{dx}=ae^{y} \\$
$\frac{dy}{e^{y}}=a (dx)\\$
$\displaystyle \int_{}^{} e^{-y} dy=a \displaystyle \int_{}^{}dx \\$
$\frac{e^{-y}}{-1}=ax+c \\$
$-e^{-y}=ax+c \\$
$0=e^{-y}+ax+c \\$
$e^{-y}+ax+c=0 \\$

$1. (ii) \; \frac{1+x^{2}}{1+y}=xy \frac{dy }{dx}\\$

$\frac{1+x^{2}}{1+y}=xy \frac{dy }{dx}\\$
$\left( \frac{1+x^{2}}{x} \right) dx= y\left( 1+y \right)dy\\$
$\displaystyle \int_{}^{}\left( \frac{1}{x}+x \right) dx = \displaystyle \int_{}^{}\left( y+y^{2} \right) dy \\$
$log(x)+\frac{x^{2}}{2}=\frac{y^{2}}{2}+\frac{y^{3}}{3}+c \\$

$2.\; Solve\; y(1-x)-x\frac{dy}{dx}=0 \\$
$y(1-x)=x\frac{dy}{dx}\\$
$\left( \frac{1-x}{x} \right)dx=\frac{dy}{y} \\$
$\displaystyle \int_{}^{}\left( \frac{1}{x}-1 \right) dx=\displaystyle \int_{}^{}\frac{dy}{y} \\$
$log\; x-x=log \;y+c\\$

$3. (i) \; Solve \;y \; dx -x\; dy=0 \\$

$\;y \; dx -x\; dy=0 \\$
$y \; dx =x\; dy \\$
$\displaystyle \int_{}^{}\frac{dx}{x} =\displaystyle \int_{}^{}\frac{dy}{y} \\$
$log \; x=log \; y+ log \; c\\$
$log \; x=log\; yc \\$

Taking exponential on both sides,

x=yc

$3. (ii) \; Solve\; \frac{dy}{dx}+e^{x}+ye^{x}=0 \\$
$\frac{dy}{dx}+e^{x}+ye^{x}=0 \\$
$\frac{dy}{dx}+e^{x} \left( 1+y \right)=0 \\$
$\frac{dy}{dx}=-e^{x}\left( 1+y \right) \\$
$\displaystyle \int_{}^{}\frac{dy}{1+y}=-\displaystyle \int_{}^{}e^{x} dx \\$
$log (1+y)=-e^{x}+c \\$

$4.. \; Solve\; cos\;x(1+cos\;y) dx-sin\;y(1+sin\;x)dy=0\\$
$cos\;x(1+cos\;y) dx-sin\;y(1+sin\;x)dy=0\\$
$cos\;x(1+cos\;y) dx=sin\;y(1+sin\;x)dy \\$
$\frac{cos\;x}{\left( 1+sin\; x \right)} dx= \frac{sin\;y}{\left( 1+cos\; y \right)} dy\\$
Multiply and divide by – on RHS

$\frac{cos\;x}{\left( 1+sin\; x \right)} dx= -\frac{-sin\;y}{\left( 1+cos\; y \right)} dy \\$
$log {\left( 1+sin\; x \right)} + log (1+cos\; y )=log \;c \\$
$Take\; t=1+sin\; x\\$
$dt=cos\;x dx\\$
$\displaystyle \int_{}^{}\frac{dt}{t} \\$
$log\;t\\$
$log\left( 1+sin\; x \right)\\$
Similarly,

$\displaystyle \int_{}^{}\frac{sin\;y}{\left( 1+cos\; y \right)} dy = log\left( 1+cos\; y \right) \\$

Taking exponential on both sides

$\left( 1+sin\;x \right) \left( 1+cos\;y \right)=c\\$

$5. \; Solve\; \left( 1-x \right)\;dy-\left( 1+y \right)\;dx=0 \\$
$\left( 1-x \right)\;dy-\left( 1+y \right)\;dx=0 \\$
$\left( 1-x \right)\;dy=\left( 1+y \right)\;dx \\$
$\displaystyle \int_{}^{}\frac{dy}{1+y}= \displaystyle \int_{}^{}\frac{dx}{1-x} \\$
$log\left( 1+y \right)=log\left( 1-x\right)+log\;c \\$
$log\left( 1+y \right)=\frac{log\left( x-1\right)}{-1}+log\;c \\$
$log\left( 1+y \right)= -log\left( 1-x \right)+log\;c \\$
$log\left( 1+y \right)+log\left( 1-x \right)=log\;c \\$
$log\left( 1-x \right)\left( 1+y \right)=log\;c \\$
Taking exponential on both sides

$\left( 1-x \right)\left( 1+y \right)=c \\$

$6. (i)\; Solve\; \frac{dy }{dx}= y \;sin2x\\$
$\displaystyle \int_{}^{}\frac{dy}{y}= \displaystyle \int_{}^{}sin2x \;dx \\$
$log\;y=-\displaystyle \frac{cos2x}{2}+c \\$
$\\$

$6.(ii)\; log\frac{dy }{dx}=ax+by \\$
Taking exponential on both sides

$\displaystyle \frac{dy}{dx}=e^{ax+by} \\$
$\displaystyle \frac{dy}{dx}=e^{ax}\;e^{by} \\$
$\displaystyle \frac{dy}{e^{by}}=e^{ax}\;dx \\$
$\displaystyle \int_{}^{}e^{-by} dy=\displaystyle \int_{}^{}e^{ax} dx \\$
$\displaystyle \frac{e^{-by}}{-b}=\frac{e^{ax}}{a}+c \\$
$0=\displaystyle \frac{e^{ax}}{a}+ \displaystyle \frac{e^{-by}}{b}+c \\$
$\displaystyle \frac{e^{ax}}{a}+\frac{e^{-by}}{b}+c=0 \\$

7. Find the curve whose gradient at any point P(x,y) on it is $\displaystyle \frac{x-a}{y-b} \$ and which passes through the origin.

$gradient\;= \displaystyle \frac{x-a}{y-b} \\$
$\displaystyle \frac{dy}{dx}=\displaystyle \frac{x-a}{y-b} \\$
$\left( y-b \right)dy=\left( x-a \right)dx \\$
$\displaystyle \int_{}^{}\left( y-b \right)dy=\displaystyle \int_{}^{}\left( x-a \right)dx \\$
$\displaystyle \frac{y^{2}}{2}-by= \displaystyle \frac{x^{2}}{2}-ax+c\Rightarrow (1) \\$
Since the curve passes through (0,0) substituting in (1)

$0-0=0-0+c \\$
$c=0 \\$
$Eq \;(1)\displaystyle \frac{y^{2}}{2}-by=\frac{x^{2}}{2}-ax \\$

$\displaystyle \frac{y^{2}-2by}{2}=\frac{x^{2}-2ax}{2} \\$
$y^{2}-2by=x^{2}-2ax \\$
$y^{2}-2by+b^{2}-b^{2}=x^{2}-2ax+a^{2}-a^{2} \\$
$\left( y-b\right)^{2}-b^{2}=\left( x-a\right)^{2}+b^{2}-a^{2} \\$

Differential Equations Exercise 4.2 Text Book Solutions

Related Posts:

Trackbacks

Leave a Reply Cancel reply

Recent

Related Posts:

Reader Interactions

Trackbacks

Leave a Reply Cancel reply

Footer

Recent

Tags