Differential Equations Exercise 4.4 Text Book Solutions

Dear Class 12 Samacheer Kalvi students, here are the Business Maths Chapter 4 Differential Equations Exercise 4.4 Text Book Solutions for your reference and study.

You can find links to the other exercises also below.

Important Formulas and Notes in Differential Equations

Text Book Solutions for Differential Equations Exercise 4.1

Text Book Solutions for Differential Equations Exercise 4.2

Text Book Solutions for Differential Equations Exercise 4.3

Text Book Solutions for Differential Equations Exercise 4.5

Text Book Solutions for Differential Equations Exercise 4.6

Exercise 4.4

$1.\; \displaystyle \frac{dy }{dx}-\displaystyle \frac{y}{x}=x\\$
$The \;given \;equation \;is \;linear \;in \;y \\$
$p=\displaystyle \frac{1}{x} ; Q=x \\$
$I.F = e^{\int P\;dx} \\$
$\displaystyle \int P\;dx= \displaystyle \int \displaystyle \frac{-1}{x}\;dx \\$
$\displaystyle \int P\;dx=-log\;x\\$
$I.F =e^{-log\;x} \\$
$I.F =e^{log\;x^{-1}} \\$
$I.F = x^{-1} \\$
$I.F =\displaystyle \frac{1}{x} \\$
$The \; general \; solution \; is \; y\; (I.F)=\displaystyle \int Q (I.F)\;dx+c\\$
$y\;\left( \displaystyle \frac{1}{x} \right) =\displaystyle \int x\; \left( \displaystyle \frac{1}{x} \right)\;dx+c \\$
$\displaystyle \frac{y}{x} = \displaystyle \int dx +c \\$
$\displaystyle \frac{y}{x} = x +c \\$

$2.\; \displaystyle \frac{dy}{dx}+y\; cos\;x = sin\;x\;cos\;x \\$
$The \; given \; equation \; is \; linear \; in \; y \\$
$P=cos\;x ; \;Q=sin\;x\;cos\;x \\$
$I.F = e^{\int P\;dx} \\$
${\nt P\;dx} =\displaystyle\int cos\;x\; dx \\$
${\int P\;dx} =sin\;x \\$
$I.F = e^{sin\;x} \\$
$The \; general \; solution \; is \; y\; (I.F)=\displaystyle \int Q (I.F)\;dx+c \\$
$y\; e^{sin\;x} = \displaystyle \int sin\;x\;cos\;x \; e^{sin\;x}\;dx +c \\$
$Put\; t= sin\;x \\$
$dt= cos\;x\;dx \\$
$y\; e^{sin\;x} = \displaystyle \int t\;\;e^{t\;dt}+c \\$
$\begin{array}{c|c} u = t & dv = e^{t}dt \\ u' = 1 & v = e^{t} \\ u'' = 0 & v_{1} = e^{t} \\ & v_{2} = e^{t} \\ \end{array}\\$
$Using \; integration \; by \; parts \\$
$\displaystyle \int u\;dv=u\;v-u'\;v_{1}+u''\;v_{2}-u'''\;v_{3}... \\$
$y\; e^{sin\;x} =t\;e^{t}-1(e^{t})+c \\$
$y\; e^{sin\;x} =e^{t}(t-1)+c \\$
$y\; e^{sin\;x} =e^{sin\;x}(sin\;x-1)+c \\$

$3.\;x\;\displaystyle \frac{dy }{dx}+2y=x^{4} \\$
$Dividing \; by \; x\\$
$\;\displaystyle \frac{dy }{dx}+\displaystyle \frac{2y}{x}=x^{3} \\$
$The \;given \;equation \;is \;linear \;in \;y \\$
$P=\displaystyle \frac{2}{x} ; \;Q=x^{3} \\$
$I.F = e^{\int P\;dx} \\$
$\displaystyle \int P\;dx=\displaystyle \int \frac{2}{x}\;dx \\$
$\displaystyle \int P\;d ={2\;log\;x} \\$
$I.F =e^{log\;x^{2}}\\$
$I.F=log\;x^{2} \\$
$The \; general \; solution \; is \; y\; (I.F)=\displaystyle \int Q (I.F)\;dx+c \\$
$y\; (x^{2})=\displaystyle \int x^{3}\; x^{2}\;dx+c \\$
$y\; x^{2}=\displaystyle \int x^{5}\; dx+c \\$
$y\; x^{2}=\displaystyle \frac{x^{6}}{6}+c \\$

$4. \displaystyle \frac{dy}{dx}+\left( \displaystyle \frac{3x^{2}}{1+x^{3}} \right)\;y=\displaystyle \frac{1+x^{2}}{1+x^{3}} \\$
$The \;given \;equation \;is \;linear \;in \;y \\$
$P=\displaystyle \frac{3x^{2}}{1+x^{3}} \\$
$Q=\displaystyle \frac{1+x^{2}}{1+x^{3}} \\$
$I.F = e^{\int P\;dx} \\$
$\displaystyle \int Pdx=\displaystyle \int \displaystyle \frac{3x^{2}}{1+x^{3}} \;dx \\$
$I.F=e^{log(1+x^{3})}\\$
$I.F=1+x^{3} \\$
$The \; general \; solution \; is \; y\; (I.F)=\displaystyle \int Q (I.F)\;dx+c \\$
$y\; (1+x^{3})=\displaystyle \int \displaystyle \frac{1+x^{2}}{1+x^{3}}\left( 1+x^{3} \right)dx+c \\$
$y\; (1+x^{3})=x+\displaystyle \frac{x^{3}}{3}+c \\$

$5.\; \displaystyle \frac{dy }{dx}+\displaystyle \frac{y}x}=x\;e^{x} \\$
$The \;given \;equation \;is \;linear \;in \;y \\$
$P=\displaystyle \frac{1}{x} \\$
$Q=x\;e^{x} \\$
$I.F = e^{\int P\;dx} \\$
$\displaystyle \int P\;dx =\displaystyle \frac{1}{x}\; dx \\$
$\displaystyle \int P\;dx =log\;x\\$
$I.F=e^{log\;x} \\$
$I.F=x \\$
$The \; general \; solution \; is \; y\; (I.F)=\displaystyle \int Q (I.F)\;dx+c \\$
$y\; (x)=\displaystyle \int x (e^{x})\;x \;dx+c \\$
$x\;y=\displaystyle \int x^{2} (e^{x})\;dx+c \\$
$\begin{array}{c|c} u = x^{2} & dv = e^{x}dx \\ u' = 2x & v = e^{x} \\ u'' = 2 & v_{1} = e^{x} \\ & v_{2} = e^{x} \\ \end{array}\\$
$Using \; integration \; by \; parts \\$
$\displaystyle \int u\;dv=u\;v-u'\;v_{1}+u''\;v_{2}-u'''\;v_{3}... \\$
$x\;y=\left( x^{2}e^{x}-2xe^{x}+2e^{e} \right)+c \\$
$x\;y= e^{x}\left( x^{2}-2x+2 \right)+c \\$

$6. \;\displaystyle \frac{dy }{dx}+y\;tan\;x=cos^{3}x \\$
$The \;given \;equation \;is \;linear \;in \;y \\$
$P=tan\;x \\$
$Q=cos^{3}x \\$
$I.F=e^ {\displaystyle \int {P\;dx}}\\$
$\displaystyle \int Pdx=\displaystyle \int tan\;x \;dx \\$
$\displaystyle \int Pdx=log\;sec\;x \\$
$I.F=e^{\displaystyle \int {log\;sec\;x} \\$

$I.F= sec\;x \\$
$The \; general \; solution \; is \; y\; (I.F)=\displaystyle \int Q (I.F)\;dx+c \\$
$y\; (sec\;x)=\displaystyle \int cos^3 x(sec\;x)\;dx+c \\$
$y\; (sec\;x)=\displaystyle \int cos^3 x\left( \displaystyle \frac{1}{cos\;x} \right)\;dx+c \\$
$y\; (sec\;x)=\displaystyle \int cos^2 x\;dx+c \\$
$y\; (sec\;x)=\displaystyle \int \displaystyle \frac{1+cos2x}{2}dx+c \\$
$y\; (sec\;x)=\displaystyle \frac{1}{2}\displaystyle \int\left( 1+cos2x \right)dx+c \\$
$y\; (sec\;x)=\displaystyle \frac{1}{2}\left( x+ \displaystyle \frac{sin2x}{2} \right)+c \\$

$7.\; If \; \displaystyle \frac{dy }{dx}+2y\;tan x=sin\;x\; and\; if\;y=0\;when\;x=\displaystyle \frac{\pi}{3}, \;express \;y\; in\; terms\; of\; x \\$
$\displaystyle \frac{dy }{dx}+2y\;tan x=sin\;x \\$
$The \;given \;equation \;is \;linear \;in \;y \\$
$P=2\;tan\;x \\$
$Q=2\;sin\;x \\$
$I.F=e^{\displaystyle \int Pdx}\\$
$\displaystyle \int Pdx=\displaystyle \int 2\;tan\;x\; dx \\$
$\displaystyle \int Pdx=2\displaystyle \int tan\;x\; dx \\$
$\displaystyle \int Pdx=2 \;log \;sec\;x \\$
$I.F= e^{\;2 \;log\;sec\;x} \\$
$I.F= e^{ \;log\;sec\;^{2}x} \\$
$I.F= sec\;^{2}x \\$
$The \; general \; solution \; is \; y\; (I.F)=\displaystyle \int Q (I.F)\;dx+c \\$
$y\; (sec^{2}x)=\displaystyle \int sin\;x (sec^{2}x)\;dx+c \\$
$y\; (sec^{2}x)=\displaystyle \int sin\;x (sec\;x)\;(sec\;x)\;dx+c \\$
$y\; (sec^{2}x)=\displaystyle \int sin\;x (\displaystyle \frac{1}{cos\;x})\;(sec\;x)\;dx+c \\$
$y\; (sec^{2}x)=\displaystyle \int tan\;x (sec\;x)\;dx+c \\$
$y\; (sec^{2}x)= sec\;x\;+c\Rightarrow (1) \\$
$Substitute\;x=\displaystyle \frac{\pi}{3}\;and\;y=o\; in\;(1) \\$
$0\; (sec^{2}\displaystyle \frac{\pi}{3})= sec\;\frac{\pi}{3}\;+c\Rightarrow (1) \\$
$0= 2\;+c \\$
$c=-2 \\$
$y\; (sec^{2}x)= sec\;x\;-2 \\$

8. Same as problem no 5.

$9.\;\text{A bank pays interest by continuous compounding, that is by treating the}$
$\text{interest rate as the instantaneous rate of change of principal. A man}$
$\text{invests Rs. 1,00,000 in the bank deposit which accures interest, 8\% }$
$\text{year compounded continuously. How much will he get after 10 years.}\\$

$Let\;A\; be\; any\; amount\;at\; any\; time\;t\\$
$\displaystyle \frac{dA }{dt}=8\%A \\$
$\displaystyle \frac{dA }{dt}=0.08A \\$
$When\; t=0\; \;A=100,000 \\$
$When\; t=10\; \;A=? \\$
$\displaystyle \frac{dA }{dt}-0.08A =0 \\$
$The \;given \;equation \;is \;linear \;in \;A \\$
$P=-0.08 \\$
$Q=0\\$
$I.F=e^{\displaystyle \int Pdt} \\$
$\displaystyle \int Pdt=\displaystyle \int -0.08 dt \\$
$\displaystyle \int Pdt= -0.08 t \\$
$I.F= e^{-0.08 t} \\$
$The \; general \; solution \; is \; A\; (I.F)=\displaystyle \int Q (I.F)\;dx+c \\$
$A\; ( e^{-0.08 t})=\displaystyle \int 0 ( e^{-0.08 t})\;dt+c \\$
$A\; ( e^{-0.08 t})=0+c \\$
$A\; ( e^{-0.08 t})=c \\$
$A=\displaystyle \frac{c}{ e^{-0.08 t}}\Rightarrow (1) \\$
$Substitute\;t=0; \; A=100000 \; in \;(1) \\$
$100000=c\;e^{0.08(0)} \\$
$c=100000 \\$
$A=\displaystyle {100000}\; e^{0.08 t}\Rightarrow (2) \\$
$Substitute \;t=10 \;in\; (2) \\$
$A=\displaystyle {100000}\; e^{0.08 (10)} \\$
$A=\displaystyle {100000}\; e^{0.8} \\$
$A=\displaystyle {100000}\; (2.2255) \\$
$A=222,550 \\$