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Differential Equations Exercise 4.3 Text Book Solutions

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Dear Class 12 Samacheer Kalvi students, here are the Business Maths Chapter 4 Differential Equations Exercise 4.3 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 4.4

Text Book Solutions for Differential Equations Exercise 4.5

Text Book Solutions for Differential Equations Exercise 4.6

Exercise 4.3

1. Solve the following homogeneous differential equations.

x\; \displaystyle \frac{dy }{dx}=x+y\\
Dividing by x

\displaystyle \frac{dy }{dx}=1+\displaystyle \frac{y}{x}\Rightarrow Eq(1)\\
Put y=vx

v=\displaystyle \frac{y}{x} \\
Differentiating w.r.t.x

\displaystyle \frac{dy }{dx}=v+x\displaystyle \frac{dv }{dx}\Rightarrow Eq (2)\\
From (1) and (2)

v+x\displaystyle \frac{dv }{dx}=1+\displaystyle \frac{y}{x}\\
v+x\displaystyle \frac{dv }{dx}=1+\displaystyle \frac{vx}{x} \\
v+x\displaystyle \frac{dv }{dx}=1+v \\
x\displaystyle \frac{dv }{dx}=1 \\
\displaystyle \int dv=\displaystyle \int\frac{dx}{x}\\
v=log\;x+log\;c \\
v=log\;xc\\
\displaystyle \frac{y}{x}=log\;xc\\
Taking exponential on both sides,

e^{\displaystyle\frac{y}{x}}=xc\\

2.\; (x-y)\displaystyle \frac{dy }{dx}=x+3y\\
\displaystyle \frac{dy }{dx}=\displaystyle \frac{x+3y}{x-y}\Rightarrow Eq.(1)\\
Put y=vx

v=\displaystyle \frac{y}{x} \\
Differentiating w.r.t.x

\displaystyle \frac{dy }{dx}=v+x\displaystyle \frac{dv }{dx}\Rightarrow Eq (2)\\
From (1) and (2)

v+x\displaystyle \frac{dv }{dx}=\displaystyle \frac{x+3y}{x-y} \\
v+x\displaystyle \frac{dv }{dx}=\displaystyle \frac{x+3vx}{x-vx}\\
v+x\displaystyle \frac{dv }{dx}=\displaystyle \frac{x(1+3v)}{x(1-v)}\\
v+x\displaystyle \frac{dv }{dx}=\displaystyle \frac{(1+3v)}{(1-v)}\\
x\displaystyle \frac{dv }{dx}=\displaystyle \frac{(1+3v)}{(1-v)}-v \\
x\displaystyle \frac{dv }{dx}=\displaystyle \frac{1+3v-v(1-v)}{1-v}\\
x\displaystyle \frac{dv }{dx}=\displaystyle \frac{1+3v-v+v^{2}}{1-v} \\
x\displaystyle \frac{dv }{dx}=\displaystyle \frac{v^{2}+2v+1}{-(v-1)}\\
x\displaystyle \frac{dv }{dx}=\displaystyle \frac{(v+1)^{2}}{-(v-1)}\\
\displaystyle \frac{(v-1)}{(v+1)^{2}} \;dv =-\displaystyle \frac{dx }{x}\\
\displaystyle \frac{(v+1)-2}{(v+1)^{2}} \;dv =-\displaystyle \frac{dx }{x}\\
\left( \displaystyle \frac{(v+1)}{(v+1)^{2}}-\displaystyle \frac{2}{(v+1)^{2}} \right) \;dv =-\displaystyle \frac{dx }{x}\\
\displaystyle \int_{}^{}\left( \displaystyle \frac{1}{v+1}-{2}{(v+1)^{-2}} \right) \;dv =-\displaystyle \int_{}^{}\frac{dx }{x}\\
log\;(v+1)-\displaystyle \frac{2(v+1)^{-1}}{-1}=-log\;x+c\\
log\;(v+1)+\displaystyle \frac{2}{v+1}=-log\;x+c\\
log\;(v+1)+log\; x=\displaystyle \frac{-2}{v+1}+c\\
log\;(v+1)\; x=\displaystyle \frac{-2}{v+1}+c\\
log\;(\frac{y}{x}+1)\; x=\displaystyle \frac{-2}{\displaystyle \frac{y}{x}+1}+c\\
log\;(\frac{y+x}{x})\; x=\displaystyle \frac{-2}{\displaystyle \frac{y+x}{x}}+c\\
log\;(x+y) =\displaystyle \frac{-2x}{{x+y}}+c\\
Taking exponential on both sides,
(x+y) =e^{\displaystyle \frac{-2x}{{x+y}}+c}\\
(x+y) =e^{\displaystyle \frac{-2x}{{x+y}}}\; e^{c}\\
(x+y) =c\; e^{\displaystyle \frac{-2x}{{x+y}}}\\

3.\; x\displaystyle \frac{dy }{dx}-y=\displaystyle \sqrt{x^{2}+y^{2}} \\
x\displaystyle \frac{dy }{dx}=y+\displaystyle \sqrt{x^{2}+y^{2}}\\
Dividing by x

\displaystyle \frac{dy }{dx}=\displaystyle \frac{y}{x}+\displaystyle\frac{\sqrt{x^{2}+y^{2}}}{x} \Rightarrow (1) \\
Put y=vx

v=\displaystyle \frac{y}{x} \\
Differentiating w.r.t.x

\displaystyle \frac{dy }{dx}=v+x\displaystyle \frac{dv }{dx}\Rightarrow Eq (2)\\
From (1) and (2)

v+x\displaystyle \frac{dv }{dx}=\displaystyle \frac{vx}{x}+\displaystyle\frac{\sqrt{x^{2}+v^{2}x^{2}}}{x} \\
v+x\displaystyle \frac{dv }{dx}=v+\displaystyle\frac{\sqrt{x^{2}(1+v^{2})}}{x} \\
v+x\displaystyle \frac{dv }{dx}=v+\displaystyle \frac{x\sqrt{(1+v^{2})}}{x}\\
v+x\displaystyle \frac{dv }{dx}=v+ \sqrt{1+v^{2}}\\
x\displaystyle \frac{dv }{dx}=\sqrt{1+v^{2}} \\
\displaystyle \int \displaystyle \frac{dv}{\sqrt{1+v^{2}}}=\displaystyle \int\displaystyle \frac{dx}{x}\\
We know that

\displaystyle \int \displaystyle \frac{dx}{\sqrt{a^{2}+x^{2}}}=log\; x+\displaystyle \sqrt{a^{2}+x^{2}}+c\\
=log\; (v+\displaystyle \sqrt{1+v^{2}})=log\;x+log\;c\\
=log\; (v+\displaystyle \sqrt{1+v^{2}})=log\;xc\\
Taking exponential on both sides,

v+\displaystyle \sqrt{1+v^{2}}= xc\\
\displaystyle \frac{y}{x}+\displaystyle \sqrt{1+\displaystyle \frac{y^{2}}{x^{2}}}= xc \\
\displaystyle \frac{y}{x}+\displaystyle \sqrt{\displaystyle \frac{x^{2}+y^{2}}{x^{2}}}= xc \\
\displaystyle \frac{y}{x}+\displaystyle \frac{\sqrt{x^{2}+y^{2}}}{x}=xc\\
\displaystyle \frac{y+\sqrt{x^{2}+y^{2}}}{x}=xc\\
\displaystyle {y+\sqrt{x^{2}+y^{2}}}=x^{2}c\\

4. \;\displaystyle \frac{d y}{dx}=\displaystyle \frac{3x-2y}{2x-3y}\\
\displaystyle \frac{d y}{dx}=\displaystyle \frac{3x-2y}{2x-3y}\Rightarrow (1) \\
Put y=vx

v=\displaystyle \frac{y}{x} \\
Differentiating w.r.t.x

\displaystyle \frac{dy }{dx}=v+x\displaystyle \frac{dv }{dx}\Rightarrow Eq (2)\\
From (1) and (2)

v+\displaystyle x\frac{dv}{dx}=\displaystyle \frac{3x-2y}{2x-3y} \\
v+\displaystyle x\frac{dv}{dx}=\displaystyle \frac{3x-2vx}{2x-3vx} v+\displaystyle x\frac{dv}{dx}=\displaystyle \frac {x(3-2v)}{x(2-3v)} \\
v+\displaystyle x\frac{dv}{dx}=\displaystyle \frac {3-2v}{2-3v} \\
\displaystyle x\frac{dv}{dx}=\displaystyle \frac {3-2v}{2-3v}-v \\
\displaystyle x\frac{dv}{dx}=\displaystyle \frac {3-2v-v(2-3v)}{2-3v} \displaystyle \\
\displaystyle x\frac{dv}{dx}=\displaystyle \frac {3-2v-2v+3v^{2}}{2-3v} \\
\displaystyle x\frac{dv}{dx}=\displaystyle \frac {3v^{2}-4v+3}{-(3v-2)}\\
\displaystyle \int \displaystyle \frac{3v-2}{3v^{2}-4v+3} dv=-\displaystyle \int \displaystyle \frac{dx}{x} \\
\displaystyle \frac{1}{2}\displaystyle \int \displaystyle \frac{6v-4}{3v^{2}-4v+3} dv=-\displaystyle \int \displaystyle \frac{dx}{x} \\
\displaystyle \frac{1}{2} log\; (3v^{2}-4v+3) =- log\; x+log\; c \\
log\; (3v^{2}-4v+3)^{\frac{1}{2}} = log\; \displaystyle \frac{c}{x}\\
Taking exponential on both sides,

(3v^{2}-4v+3)^{\frac{1}{2}} = \displaystyle \frac{c}{x} \\
Squaring on both sides

3v^{2}-4v+3 = \displaystyle \frac{c^{2}}{x^{2}} \\
3\displaystyle \frac{y^{2}}{x^{2}}-4\displaystyle \frac{y}{x}+3=\displaystyle \frac{c^{2}}{x^{2}}\\
\displaystyle \frac{3y^{2}-4xy+3x^{2}}{x^{2}}=\displaystyle \frac{c^{2}}{x^{2}} \\
3y^{2}-4xy+3x^{2}=c^{2}\\

5. (y^{2}-2xy)\; dx=(x^{2}-2xy)\;dy \\
\displaystyle \frac{y^{2}-2xy}{x^{2}-2xy}=\displaystyle \frac{dy}{dx} \\
\displaystyle \frac{dy}{dx}=\displaystyle \frac{y^{2}-2xy}{x^{2}-2xy}\Rightarrow (1) \\
Put y=vx

v=\displaystyle \frac{y}{x} \\
Differentiating w.r.t.x

\displaystyle \frac{dy }{dx}=v+x\displaystyle \frac{dv }{dx}\Rightarrow Eq (2)\\
From (1) and (2)

v+x\displaystyle \frac{dv}{dx}=\displaystyle \frac{v^{2}x^{2}-2x(vx)}{x^{2}-2x(vx)} \\
v+x\displaystyle \frac{dv}{dx}=\displaystyle \frac{v^{2}x^{2}-2vx^{2}}{x^{2}-2vx^{2}} \\
v+x\displaystyle \frac{dv}{dx}=\displaystyle \frac{x^{2}(v^{2}-2v)}{x^{2}(1-2v)} \\
v+x\displaystyle \frac{dv}{dx}=\displaystyle \frac{v^{2}-2v}{1-2v} \\
x\displaystyle \frac{dv}{dx}=\displaystyle \frac{v^{2}-2v}{1-2v}-v \\
x\displaystyle \frac{dv}{dx}=\displaystyle \frac{v^{2}-2v-v(1-2v)}{1-2v} \\
x\displaystyle \frac{dv}{dx}=\displaystyle \frac{v^{2}-2v-v+2v^{2}}{1-2v} \\
x\displaystyle \frac{dv}{dx}=\displaystyle \frac{3v^{2}-3v}{1-2v} \\
x\displaystyle \frac{dv}{dx}=\displaystyle \frac{3(v^{2}-v)}{-(2v-1)} \\
\displaystyle \int \displaystyle \frac{2v-1}{v^{2}-v} \;dv=-3\displaystyle \int\frac{dx }{x} \\
log\; (v^{2}-v)=-3\;log\;x+log\; c \\
log\; (v^{2}-v)=log\;x^{-3}+log\; c \\
log\; (v^{2}-v)=log\;x^{-3}c \\
Taking exponential on both sides,

v^{2}-v=x^{-3}c \\
\displaystyle \frac{y^{2}}{x^{2}}-\displaystyle \frac{y}{x}=\displaystyle \frac{c}{x^{3}} \\
\displaystyle \frac{y^{2}-xy}{x^{2}}=\displaystyle \frac{c}{x^{3}} \\
y^{2}-xy=\displaystyle \frac{c}{x} \\
x (y^{2}-xy)=c \\

6. The slope of the tangent to a curve at any point (x, y) on it is given by (y^{3}-2yx^{2})dx+(2xy^{2}-x^{3})dy and the curve passes through (1, 2). \; Find the equation of the curve.

Let (y^{3}-2yx^{2})dx+(2xy^{2}-x^{3})dy=0 \\
(2xy^{2}-x^{3})dy=-(y^{3}-2yx^{2})dx \\
\displaystyle \frac{dy}{dx}=-\displaystyle \frac{y^{3}-2yx^{2}}{2xy^{2}-x^{3}}\Rightarrow (1) \\
Put y=vx

v=\displaystyle \frac{y}{x} \\
Differentiating w.r.t.x

\displaystyle \frac{dy }{dx}=v+x\displaystyle \frac{dv }{dx}\Rightarrow Eq (2)\\
From (1) and (2)

v+x\;\displaystyle \frac{dv}{dx}=-\displaystyle \frac{y^{3}-2yx^{2}}{2xy^{2}-x^{3}} \\
v+x\;\displaystyle \frac{dv}{dx}=-\displaystyle \frac{v^{3}x^{3}-2(vx)x^{2}}{2x(vx)^{2}-x^{3}} \\
v+x\;\displaystyle \frac{dv}{dx}=-\displaystyle \frac{v^{3}x^{3}-2vx^{3}}{2v^{2}x^{3}-x^{3}} \\
v+x\;\displaystyle \frac{dv}{dx}=-\displaystyle \frac{x^{3}(v^{3}-2v)}{x^{3}(2v^{2}-1)} \\
x\;\displaystyle \frac{dv}{dx}=\displaystyle \frac{-v^{3}+2v}{2v^{2}-1}-v \\
x\;\displaystyle \frac{dv}{dx}=\displaystyle \frac{-v^{3}+2v-v(2v^{2}-1)}{2v^{2}-1} \\
x\;\displaystyle \frac{dv}{dx}=\displaystyle \frac{-v^{3}+2v-2v^{3}+v}{2v^{2}-1} \\
x\;\displaystyle \frac{dv}{dx}=\displaystyle \frac{-3v^{3}+3v}{2v^{2}-1} \\
x\;\displaystyle \frac{dv}{dx}=\displaystyle \frac{-3(v^{3}-v)}{2v^{2}-1} \\
\displaystyle \int\displaystyle \frac{2v^{2}-1}{v^{3}-v}dv =-3\displaystyle \int\displaystyle \frac{dx}{x}\Rightarrow (3) \\
Using partial fractions

\displaystyle \frac{2v^{2}-1}{v^{3}-v}= \displaystyle \frac{2v^{2}-1}{v(v^{2}-1)}\\
\displaystyle \frac{2v^{2}-1}{v^{3}-v}= \displaystyle \frac{2v^{2}-1}{v(v+1)(v-1)}\\
\displaystyle \frac{2v^{2}-1}{v^{3}-v}= \displaystyle \frac{A}{v}+\displaystyle \frac{B}{v+1}+\displaystyle \frac{C}{v-1}\Rightarrow (4) \\
\displaystyle \frac{2v^{2}-1}{v^{3}-v}= \displaystyle \frac{A(v+1)(v-1)+B(v)(v-1)+C(v)(v+1)}{(v-1)(v+1)(v)} \\
\displaystyle \frac{2v^{2}-1}{v^{3}-v}= \displaystyle \frac{A(v+1)(v-1)+B(v)(v-1)+C(v)(v+1)}{v^{3}-v} \\
2v^{2}-1= \displaystyle A(v+1)(v-1)+B(v)(v-1)+C(v)(v+1) \\
Put v=0

-1= A(1)(-1)+0+0 \\
-1= - A \\
A=1\\
Put v=1

2-1= 0+0+C(1)(2) \\
1= 2C \\
C=\displaystyle \frac{1}{2} \\
Put v=-1

2-1=0+B(-1)(-2)+0 \\
1-2B \\
B=\displaystyle \frac{1}{2} \\
Substituting the values of A,B and C in (4)

\displaystyle \frac{2v^{2}-1}{v^{3}-v}= \displaystyle \frac{1}{v}+\displaystyle \frac{\frac{1}{2}}{v+1}+\displaystyle \frac{\frac{1}{2}}{v-1}\Rightarrow (4) \\
\displaystyle \frac{2v^{2}-1}{v^{3}-v}= \displaystyle \frac{1}{v}+\displaystyle \frac{{1}}{2(v+1)}+\displaystyle \frac{{1}}{2(v-1)} \\
Substituting in (3)

\left[ \displaystyle \int\displaystyle \frac{1}{v}+\displaystyle \frac{{1}}{2(v+1)}+\displaystyle \frac{{1}}{2(v-1)} \right]dv=-3\displaystyle \displaystyle \int\frac{dx}{x}\\
log\; v+\displaystyle \frac{1}{2}log\; (v+1)+\displaystyle \frac{1}{2}log\; (v-1)=-3\;log\;x \\
log\; v+log\; (v+1)^{\frac{1}{2}}+log\; (v-1)^{\frac{1}{2}}=log\;x^{-3}+ log\; c \\
log\; v \; (v+1)^{\frac{1}{2}} (v-1)^{\frac{1}{2}}=log\;x^{-3} c \\
Taking exponential on both sides,

v \; (v+1)^{\frac{1}{2}} (v-1)^{\frac{1}{2}}=x^{-3} c \\
v \; \displaystyle \sqrt{(v+1)(v-1) }=\displaystyle \frac{c}{x^{3}} \\
Squaring on both sides

v^{2} \; (v+1)(v-1) =\displaystyle \frac{c^{2}}{x^{6}} \\
\displaystyle \frac{y^{2}}{x^{2}} \Big(\displaystyle \frac{y}{x}+1\Big)\Big(\displaystyle \frac{y}{x}-1\Big)=\displaystyle \frac{c^{2}}{x^{6}} \\
\displaystyle \frac{y^{2}}{x^{2}} \Big(\displaystyle \frac{y^{2}}{x^{2}}-1\Big)=\displaystyle \frac{c^{2}}{x^{6}} \\
\displaystyle \frac{y^{2}}{x^{2}}\left( \displaystyle \frac{y^{2}-x^{2}}{x^{2}} \right)=\displaystyle \frac{c^{2}}{x^{6}} \\
\left( \displaystyle \frac{y^{2}\;(y^{2}-x^{2})}{x^{4}} \right)=\displaystyle \frac{c^{2}}{x^{6}} \\
{y^{2}\;(y^{2}-x^{2})}=\displaystyle \frac{c^{2}}{x^{2}} \\
{x^{2}\; y^{2}\;(y^{2}-x^{2})}=c^{2} \Rightarrow (5) \\
Since the curve passes through(1,2)

{(1^{2})\; (2^{2})\;(2^{2}-1^{2})}=c^{2} \\
{(4)\;\;(3)}=c^{2} \\
c^{2}=12\\
Substituting in (5)

{x^{2}\; y^{2}\;(y^{2}-x^{2})}=12\\

7. An electric manufacturing company makes small household switches. The company estimates the marginal revenue function for these switches to be (x^2 + y^22 )dy = xydx where x represents the number of units (in thousands). What is the total revenue function?

\left( x^{2}+y^{2} \right)dy =xy \;dx\\
\displaystyle \frac{dy }{dx}=\displaystyle \frac{xy}{x^{2}+y^{2}}\Rightarrow (1) \\
Put y=vx

v=\displaystyle \frac{y}{x} \\
Differentiating w.r.t.x

\displaystyle \frac{dy }{dx}=v+x\displaystyle \frac{dv }{dx}\Rightarrow Eq (2)\\
From (1) and (2)

v+ x\displaystyle \frac{dv }{dx}=\displaystyle \frac{xy}{x^{2}+y^{2}} \\
v+ x\displaystyle \frac{dv }{dx}=\displaystyle \frac{x(vx)}{x^{2}+v^{2}x^{2}} \\
v+ x\displaystyle \frac{dv }{dx}=\displaystyle \frac{vx^{2}}{x^{2}(1+v^{2})} \\
v+ x\displaystyle \frac{dv }{dx}=\displaystyle \frac{v}{(1+v^{2})} \\
x\displaystyle \frac{dv }{dx}=\displaystyle \frac{v}{(1+v^{2})}-v \\
x\displaystyle \frac{dv }{dx}=\displaystyle \frac{v-v(1+v^{2})}{(1+v^{2})} \\
x\displaystyle \frac{dv }{dx}=\displaystyle \frac{v-v-v^{3}}{(1+v^{2})} \\
x\displaystyle \frac{dv }{dx}=\displaystyle \frac{-v^{3}}{(1+v^{2})} \\
\displaystyle \int\displaystyle \frac{1+v^{2}}{v^{3}}=-\displaystyle \frac{dx }{x} \\
\displaystyle \int\left( \displaystyle \frac{1}{v^{3}} +\frac{v^{2}}{v^{3}} \right)dv =-\displaystyle \int \displaystyle \frac{dx }{x} \\
\displaystyle \int\left( v^{-3}+\displaystyle \frac{1}{v} \right)dv=-\displaystyle \int \displaystyle \frac{dx}{x} \\
\displaystyle \frac{v^{-2}}{-2}+log\;v=-log\;x+log\;c \\
\displaystyle \frac{-1}{2v^{2}}+log\;v=log\displaystyle \frac{c}{x} \\
\displaystyle \frac{-1}{2\displaystyle \frac{y^{2}}{x^{2}}}+log\frac{y}{x}=log\displaystyle \frac{c}{x} \\
\displaystyle \frac{-x^{2}}{2\displaystyle {y^{2}}}+log\frac{y}{x}=log\displaystyle \frac{c}{x} \\
log\displaystyle \frac{y}{x}-log\displaystyle \frac{c}{x}= \displaystyle \frac{x^{2}}{2\displaystyle {y^{2}}} \\
log\frac{\frac{y}{x}}{\frac{c}{x}}= \displaystyle \frac{x^{2}}{2\displaystyle {y^{2}}} \\
log\displaystyle \frac{y}{c}= \displaystyle \frac{x^{2}}{2\displaystyle {y^{2}}} \\
Taking exponential on both sides,

\displaystyle \frac{y}{c}= e^{\displaystyle \frac{x^{2}}{2\displaystyle {y^{2}}}} \\
y= c\; e^{\displaystyle \frac{x^{2}}{2\displaystyle {y^{2}}}} \\

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