\(\frac { dy }{ dx } \)+y tanx=cos³x
இது \(\frac{d y}{d x}+P y=\mathrm{Q}\) என்ற அமைப்பில் உள்ளது
இங்கு, \(\mathrm{P} =\tan x, \mathrm{Q}=\cos ^3 x \)
\(\int P d x =\int \tan x d x=\log \sec x \)
தொ.கா.
\(=e^{\int P(x)}=\mathrm{e}^{\text {log sec } x}=\sec x\)
தீர்வு :
 \( y e^{\int P d x}=\int Q e^{\int Pd x} d x+c \)
\( y \sec x=\int \cos ^3 x \cdot \sec x d x+c \)
\( =\int \cos ^3 x \frac{1}{\cos x} d x+c \)
\( =\int \cos ^2 x d x+c \)
\( =\int \frac{1+\cos 2 x}{2} d x+c \)
\( y \sec x=\frac{1}{2}\left(x+\frac{\sin 2 x}{2}\right)+c \)

\(\left(x^2+y^2\right) \mathrm{dy} =x y \mathrm{~d} x \)
\(\frac{d y}{d x} =\frac{x y}{x^2+y^2} \)
y=vx மற்றும் \(\frac { dy }{ dx } =v+x\frac { dv }{ dx } \) என (1) –ல் பிரதியிட,
\(v+x \frac{d v}{d x} =\frac{v x^2}{x^2+v^2 x^2}=\frac{v}{1+v^2} \)
\(x \frac{d v}{d x} =\frac{v}{1+v^2}-v=\frac{v-v-v^3}{1+v^2} \)
\(\int \frac{\left(1+v^2\right) \mathrm{d} v}{v^3} =-\int \frac{d x}{x} \)
\(\int \frac{d v}{v^3}+\int \frac{v^2 d v}{v^3} =-\log x+\log \mathrm{c} \)
\(\int v^{-3} d v+\int \frac{d v}{v} =\log \frac{c}{x} \)
\(\frac{v^{-2}}{-2} =\log \frac{c}{x}-\log v \)
\(-\frac{1}{2 v^2} =\log \frac{c}{x v} \)
\(-\frac{1}{2 y^2 / x^2} =\log \frac{c}{x} \times \frac{x}{y} \)
\(\frac{-x^2}{2 y^2} =\log \frac{c}{y} \)
\(\frac{c}{y} =e^{-x^3 / 2 y^2} \Rightarrow \mathrm{y}=c e^{x^2 / 2 y^2} \)

\(\left(\mathrm{y}^3-2 \mathrm{y} x^2\right) \mathrm{d} x+\left(2 x y^2-x^3\right) \mathrm{dy}=0 \)
\(\left(2 x y^2-x^3\right) \mathrm{dy}=-\left(\mathrm{y}^3-2 \mathrm{y} x^2\right) \mathrm{d} x\)
\(\frac{d y}{d x}=-\frac{\left(y^3-2 y x^2\right)}{2 x y^2-x^3} \)
y=vx மற்றும் \(\frac { dy }{ dx } =v+x\frac { dv }{ dx } \) என (1) –ல் பிரதியிட,
\( v+x \frac{d v}{d x}=\frac{-\left(v^3 x^3-2 v x^3\right)}{2 v^2 x^3-x^3} \)
\( x \frac{d v}{d x}=\frac{-\left(v^3-2 v\right)}{2 v^2-1}-v \)
\(=\frac{-v^3+2 v-2 v^3+v}{2 v^2-1} \)
\(x \frac{d v}{d x}=\frac{-3 v^3+3 v}{2 v^2-1}=\frac{-3\left(v^3-v\right)}{2 v^2-1} \)
\(\int \frac{\left(2 v^2-1\right) d v}{v^3-v}=-3 \int \frac{d x}{x} \) ...(2)
\( \frac{2 v^2-1}{v^3-v}=\frac{2 v^2-1}{v\left(v^2-1\right)} \) எனக் கருதுக 
\( \frac{2 v^2-1}{v(v+1)(v-1)}=\frac{A}{v}+\frac{B}{v+1}+\frac{C}{v-1} \)
\( 2 \mathrm{v}^2-1=\mathrm{A}(\mathrm{v}+1)(\mathrm{v}-1) +B v(v-1)+C v(v+1) \)
\( \mathrm{v}=1 \) எனில் \(1=2 C \Rightarrow C=1 / 2 \)
\( v=-1 \) எனில் \(1=2 B \Rightarrow B=1 / 2 \)
v = 0 எனில் \( -1=-A \Rightarrow A=1 \)
\(\frac{2 v^2-1}{v^3-v}=\frac{1}{v}+\frac{1}{2(v+1)}+\frac{1}{2(v-1)}\)
(2) - ல் பிரதியிட,
\( \int \frac{d v}{v}+\frac{1}{2} \int \frac{d v}{v+1}+\frac{1}{2} \int \frac{d v}{v-1}=-3 \int \frac{d x}{x} \)
\( \log \mathrm{v}+\frac{1}{2}[\log (\mathrm{v}+1)+\log (\mathrm{v}-1)] \)
\(=-3 \log x+\log c \)
\( \log \mathrm{v}+\log \left(\mathrm{v}^2-1\right)^{1 / 2}=\log \frac{1}{x^3}+\log \mathrm{c} \)
\( \log v \sqrt{v^2-1}=\log \frac{c}{x^3} \)
\( \frac{y}{x} \sqrt{\frac{y^2}{x^2}-1}=\frac{c}{x^3} \)
\( \frac{y \sqrt{y^2-x^2}}{x^2}=\frac{c}{x^3} \)
\( x y \sqrt{y^2-x^2}=\mathrm{c}\)
இது (1,2) வழிச் செல்வதால் 
\(2 \sqrt{4-1} = c \)
\(c = 2 \sqrt{3} \)
தேவையான சமன்பாடு \( x y \sqrt{y^2-x^2}=2 \sqrt{3} \)

x-y\(\frac { dx }{ dy } =a\left( { x }^{ 2 }+\frac { dx }{ dy } \right) \)
\(x-y\frac { dx }{ dy } =ax^{ 2 }+a\frac { dx }{ dy } \)
x-ax²=\(a\frac { dx }{ dy } +y\frac { dx }{ dy } \)
x(1-ax)=(a+y)\(\frac { dx }{ dy } \)
மாறிகளை பிரிக்க, நாம் பெறுவது
\(\frac { dx }{ x(1-ax) } =\frac { dy }{ a+y } \)
\(\left( \frac { a }{ 1-ax } +\frac { 1 }{ x } \right) dx=\frac { dy }{ a+y } \)
தொகையிட, \(\int { \left( \frac { a }{ 1-ax } +\frac { 1 }{ x } \right) dx } =\int { \frac { dy }{ a+y } } \)
-log(1-ax)+logx=log(a+y)+logc
\(\log\left( \frac { x }{ 1-ax } \right) \)=log(c(a+y))
\(\left( \frac { x }{ 1-ax } \right) \)=c(a+y)
x = (1-ax)(a+y)c என்பது தேவையான தீர்வாகும்.

3e^xtan ydx+(1+e^x)sec²ydy=0
3e^xtan ydx=-(1+e^x)sec²y dy
\(\frac { { 3e }^{ x } }{ 1+e^{ x } } dx=\frac { \sec^{ 2 }y }{ \tan y } dy\)
தொகையிட நாம் பெறுவது \(3\int { \frac { { e }^{ x } }{ 1+e^{ x } } } dx=-\int { \frac { sec^{ 2 }y }{ \tan y } } dy\)
3log(1+e^x)=-log tan y+logc   \(\left[ \because \int { \frac { f'(x) }{ f(x) } dx } =\log f(x) \right] \)
log(1+e^x)³+log tan y=log c
log[(1+e^x)3tan y]=log c
(1+e^x)³ tan y=c         .....(1)
தரவு: y(0)=\(\frac { \pi }{ 4 } \)  (அதாவது) x=0 எனில் y=\(\frac { \pi }{ 4 } \)
(1) ⇒ (1+e⁰)³ tan\(\frac { \pi }{ 4 } \)=c
2³(1)=c
⇒ c=8
தேவையான தீர்வு (1+e^x)³tan y=8