Ordinary Differential Equations and Exact Equations

Module 3

Ordinary differential equations of first order
3.1 Linear and Bernoulli’s differential equations

Introduction:

1. Linear differential equation in 𝒚:
𝑑𝑦
This is of the form 𝑑𝑥 + 𝑃𝑦 = 𝑄, where 𝑃 and 𝑄 are functions of 𝑥 alone.

General solution is 𝑦. 𝐼𝐹 = ∫ 𝑄. 𝐼𝐹 𝑑𝑥 + 𝑐, where 𝐼𝐹 = 𝑒 ∫ 𝑃𝑑𝑥 .

2. Linear differential equation in 𝒙:
𝑑𝑥
This is of the form + 𝑃𝑥 = 𝑄, where 𝑃 and 𝑄 are functions of 𝑦 alone.
𝑑𝑦

General solution is 𝑥. 𝐼𝐹 = ∫ 𝑄. 𝐼𝐹 𝑑𝑦 + 𝑐, where 𝐼𝐹 = 𝑒 ∫ 𝑃𝑑𝑦 .

3. Bernoulli’s differential equation in 𝒚:
𝑑𝑦
This is of the form 𝑑𝑥 + 𝑃𝑦 = 𝑄𝑦 𝑛 , where 𝑃 and 𝑄 are functions of 𝑥 alone.
𝑑𝑦
Dividing this equation by 𝑦 𝑛 , 𝑦 −𝑛 + 𝑃𝑦 1−𝑛 = 𝑄 ---- (1)
𝑑𝑥

𝑑𝑦 𝑑𝑡
Put 𝑦1−𝑛 = 𝑡, then (1 − 𝑛)𝑦 −𝑛 = 𝑑𝑥
𝑑𝑥

1 𝑑𝑡
Equation (1) becomes 1−𝑛 + 𝑃𝑡 = 𝑄
𝑑𝑥

𝑑𝑡
Reduced linear differential equation is 𝑑𝑥 + (1 − 𝑛)𝑃𝑡 = (1 − 𝑛)𝑄

4. Bernoulli’s differential equation in 𝒙:
𝑑𝑦
This is of the form 𝑑𝑥 + 𝑃𝑥 = 𝑄𝑥 𝑛 , where 𝑃 and 𝑄 are functions of 𝑦 alone.
𝑑𝑦
Dividing this equation by 𝑥 𝑛 , 𝑥 −𝑛 + 𝑃𝑥1−𝑛 = 𝑄 ---- (1)
𝑑𝑥

𝑑𝑥 𝑑𝑡
Put 𝑥1−𝑛 = 𝑡, then (1 − 𝑛)𝑥 −𝑛 = 𝑑𝑦
𝑑𝑦

1 𝑑𝑡
Equation (1) becomes 1−𝑛 + 𝑃𝑡 = 𝑄
𝑑𝑦

𝑑𝑡
Reduced linear differential equation is 𝑑𝑦 + (1 − 𝑛)𝑃𝑡 = (1 − 𝑛)𝑄




Dr. Narasimhan G, RNSIT 1

,Problems:

𝒅𝒚
1. Solve 𝒅𝒙 + 𝒚 𝒄𝒐𝒕 𝒙 = 𝒄𝒐𝒔 𝒙

This is a linear L.D.E in 𝑦 with 𝑃 = cot 𝑥 , 𝑄 = cos 𝑥
𝐼𝐹 = 𝑒 ∫ 𝑃𝑑𝑥 = 𝑒 ∫ cot 𝑥𝑑𝑥 = sin 𝑥
General solution is 𝑦. 𝐼𝐹 = ∫ 𝑄. 𝐼𝐹 𝑑𝑥 + 𝑐
𝑦. sin 𝑥 = ∫ cos 𝑥 . sin 𝑥 𝑑𝑥 + 𝑐
1
𝑦 sin 𝑥 = 2 ∫ sin 2𝑥 𝑑𝑥 + 𝑐
1
𝑦 sin 𝑥 = − 4 cos 2𝑥 + 𝑐


𝒅𝒚
2. Solve 𝒅𝒙 + 𝒚 𝒕𝒂𝒏 𝒙 = 𝒚𝟑 𝒔𝒆𝒄 𝒙

Step 1: Reduce it to an LDE
Divide by 𝑦 3 on both sides,
1 𝑑𝑦 1
+ 𝑦 2 tan 𝑥 = sec 𝑥 ---- (1)
𝑦 3 𝑑𝑥
1 2 𝑑𝑦 𝑑𝑡
If 𝑦 2 = 𝑡 then − 𝑦 3 𝑑𝑥 = 𝑑𝑥
1 1 𝑑𝑦 1 𝑑𝑡
Put 𝑦 2 = 𝑡, = − 2 𝑑𝑥 in (1)
𝑦 3 𝑑𝑥
1 𝑑𝑡
− 2 𝑑𝑥 + 𝑡 tan 𝑥 = sec 𝑥

Multiply by -2 on both sides,
𝑑𝑡
− 2𝑡 tan 𝑥 = −2 sec 𝑥
𝑑𝑥

This is an LDE in 𝑡 with 𝑃 = −2 tan 𝑥 , 𝑄 = −2 sec 𝑥
Step 2: Solve reduced LDE
𝐼𝐹 = 𝑒 ∫ −2 tan 𝑥𝑑𝑥 = 𝑒 −2 log sec 𝑥 = cos 2 𝑥
General solution is 𝑡. 𝐼𝐹 = ∫ 𝑄. 𝐼𝐹 𝑑𝑥 + 𝑐
𝑡 cos2 𝑥 = ∫ −2 sec 𝑥 cos2 𝑥 𝑑𝑥 + 𝑐
𝑡 cos2 𝑥 = −2 sin 𝑥 + 𝑐
1
(𝑦 2 ) cos 2 𝑥 = −2 sin 𝑥 + 𝑐




Dr. Narasimhan G, RNSIT 2

, 𝒅𝒚 𝒚
3. Solve + = 𝒚𝟐 𝒙 (May 22)
𝒅𝒙 𝒙


Step 1: Reduce it to an LDE

Divide by 𝑦 2 on both sides,

1 𝑑𝑦 1 1
+ 𝑥 (𝑦) = 𝑥 ---- (1)
𝑦 2 𝑑𝑥

1 1 𝑑𝑦 𝑑𝑡
If 𝑦 = 𝑡 then − 𝑦 2 𝑑𝑥 = 𝑑𝑥

1 1 𝑑𝑦 𝑑𝑡
Put 𝑦 = 𝑡, = − 𝑑𝑥 in (1)
𝑦2 𝑑𝑥

𝑑𝑡 𝑡
− 𝑑𝑥 + 𝑥 = 𝑥

Multiply by -1 on both sides,

𝑑𝑡 𝑡
− 𝑥 = −𝑥
𝑑𝑥


This is an LDE in 𝑡 with 𝑃 = −1/𝑥, 𝑄 = −𝑥

Step 2: Solve reduced LDE
1
1
𝐼𝐹 = 𝑒 ∫ −𝑥 𝑑𝑥 = 𝑒 − log 𝑥 = 𝑥

General solution is 𝑡. 𝐼𝐹 = ∫ 𝑄. 𝐼𝐹 𝑑𝑥 + 𝑐

1 1
𝑡 𝑥 = ∫ − 𝑥 𝑥 𝑑𝑥 + 𝑐

1
𝑡 𝑥 = −𝑥 + 𝑐

1
= −𝑥 + 𝑐
𝑥𝑦




Dr. Narasimhan G, RNSIT 3

, 𝒅𝒚
4. 𝒕𝒂𝒏 𝒚 + 𝒕𝒂𝒏 𝒙 = 𝒄𝒐𝒔 𝒚 𝒄𝒐𝒔𝟐 𝒙
𝒅𝒙


Step 1: Reduce it to an LDE

Divide by cos 𝑦 on both sides,

𝑑𝑦
sec 𝑦 tan 𝑦 𝑑𝑥 + sec 𝑦 tan 𝑥 = cos 2 𝑥 ---- (1)

𝑑𝑦 𝑑𝑡
If sec 𝑦 = 𝑡 then sec 𝑦 tan 𝑦 𝑑𝑥 = 𝑑𝑥

𝑑𝑡
(1) ⇒ 𝑑𝑥 + 𝑡 tan 𝑥 = cos2 𝑥

This is an LDE in 𝑡 with 𝑃 = tan 𝑥 , 𝑄 = cos 2 𝑥

Step 2: Solve reduced LDE

𝐼𝐹 = 𝑒 ∫ 𝑃 𝑑𝑥 = 𝑒 ∫ tan 𝑥 𝑑𝑥 = sec 𝑥

General solution is 𝑡. 𝐼𝐹 = ∫ 𝑄. 𝐼𝐹 𝑑𝑥 + 𝑐

𝑡 sec 𝑥 = ∫ cos 2 𝑥 sec 𝑥 𝑑𝑥 + 𝑐

sec 𝑦 sec 𝑥 = sin 𝑥 + 𝑐




Dr. Narasimhan G, RNSIT 4