Differential Equations of Order
One
2.4 Linear Equation of Order One
2.4 Linear Equation of Order One
• An equation that is linear and of order one in the independent
𝑑𝑦
variable 𝑦 must be of the form 𝐴 𝑥 + 𝐵 𝑥 𝑦 = 𝐶 𝑥 (1)
𝑑𝑥
• By dividing each member of equation (1) by 𝐴 𝑥 , we obtain
𝑑𝑦
+ 𝑃 𝑥 𝑦 = 𝑄 𝑥 2 ,which we choose as the standard from for
𝑑𝑥
the linear equation of order one.
• Suppose that there exists for equation 2 a positive integrating
factor 𝑣 𝑥 > 0, a function of 𝑥 alone. Then,
𝑑𝑦
𝑣 𝑥 + 𝑃 𝑥 𝑦 = 𝑣(𝑥) ∙ 𝑄(𝑥) 3
𝑑𝑥
must be an exact equation.
• Equation 3 is easily put into the form 𝑀 𝑑𝑥 + 𝑁 𝑑𝑦 = 0 with
𝑀 = 𝑣𝑃𝑦 − 𝑣𝑄 and 𝑁 = 𝑣, in which 𝑣, 𝑃, and 𝑄 are functions of 𝑥
alone.
• Since equation 3 is to be exact, it follows that
𝜕𝑀 𝜕𝑁
=
𝜕𝑦 𝜕𝑥
𝑑𝑣
⟹ 𝑣𝑃 =
𝑑𝑥
𝑑𝑣
⟹ = 𝑃 𝑑𝑥
𝑣
⟹ ln 𝑣 = 𝑃 𝑑𝑥
⟹ 𝑣 = 𝑒 𝑃 𝑑𝑥
• Let us multiply 2 by the integrating factor and we get,
𝑃 𝑑𝑥 𝑑𝑦
𝑒 + 𝑃 𝑥 𝑦 = 𝑄(𝑥)
𝑑𝑥
𝑃 𝑑𝑥 𝑑𝑦 𝑃 𝑑𝑥 𝑃 𝑑𝑥
⟹ 𝑒 + 𝑃 (𝑒 )𝑦 = 𝑄𝑒 4
𝑑𝑥
• The left member of 4 is the derivative of the product 𝑦 𝑒 𝑃 𝑑𝑥
and the right member of 4 is a function of 𝑥 alone. Hence, the
equation 4 is exact.
We summarize the steps involved in solving a linear equation of order
one:
𝑑𝑦
a. Put the equation into standard form: + 𝑃𝑦 = 𝑄.
𝑑𝑥
b. Obtain the integrating factor 𝑒 𝑃 𝑑𝑥 .
c. Multiply both sides of the equation (in standard form) by the
integrating factor.
d. Solve the resultant exact equation.
Note in integrating the exact equation that the integral of the left
member is always the product of the dependent variable and the
integrating factor used.
Solve the following differential equations.
1. 2 𝑦 − 4𝑥 2 𝑑𝑥 + 𝑥 𝑑𝑦 = 0
Solution:
Note that the equation is linear in 𝑦. Then writing it in standard form
will give us,
𝑑𝑦 2
+ 𝑦 = 8𝑥 1 when 𝑥 ≠ 0.
𝑑𝑥 𝑥
2
Now, 𝑃 = and so the integrating factor is;
𝑥
2 𝑑𝑥 2
𝑃 𝑑𝑥 𝑑𝑥 2
𝑒 =𝑒 𝑥 =𝑒 =𝑥 𝑒 2𝑙𝑛 𝑥
= 𝑒 𝑙𝑛𝑥 = 𝑥2
Next multiply the integrating factor to 1 ,
𝑑𝑦 2
𝑥2 + 𝑦 = 8𝑥
𝑑𝑥 𝑥
2 𝑑𝑦
⟹ 𝑥 + 2𝑥𝑦 = 8𝑥 3
𝑑𝑥
𝑑
⟹ 𝑥 2 𝑦 = 8𝑥 3
𝑑𝑥
𝑑
⟹ 𝑥 2 𝑦 = 8𝑥 3
𝑑𝑥
⟹ 𝑥 𝑦 = 2𝑥 4 + 𝑐
2 ∎
2. 𝑦 𝑑𝑥 + 3𝑥 − 𝑥𝑦 + 2 𝑑𝑦 = 0
Solution:
The equation is not linear in 𝑦, since the product of 𝑦 and 𝑑𝑦 occurs in
the equation. However, it is linear in 𝑥 and so the standard form will be
𝑑𝑥
+ 𝑃 𝑦 𝑥 = 𝑄(𝑦).Hence, we have
𝑑𝑦
𝑑𝑥 3 −2
+ −1 𝑥 = 𝑦≠0 1
𝑑𝑦 𝑦 𝑦
Now, we can solve the integrating factor
3
𝑃𝑑𝑦 −1 𝑑𝑦
𝑒 =𝑒 𝑦
= 𝑒 3 ln 𝑦 −𝑦
= 𝑒 3 ln 𝑦 ∙ 𝑒 −𝑦
ln 𝑦 3 −𝑦
= 𝑒 𝑒
= 𝑦 3 𝑒 −𝑦
Therefore, by multiplying the integrating factor to 1 we have
𝑑𝑥 3 −2
𝑦 3 𝑒 −𝑦 + −1 𝑥 =
𝑑𝑦 𝑦 𝑦
⟹ 𝑦 3 𝑒 −𝑦 𝑑𝑥 + 𝑦 2 3 − 𝑦 𝑒 −𝑦 𝑥 𝑑𝑦 = −2𝑦 2 𝑒 −𝑦 𝑑𝑦
⟹ 𝑦 3 𝑒 −𝑦 𝑑𝑥 + 𝑦 2 3 − 𝑦 𝑒 −𝑦 𝑥 𝑑𝑦 = −2𝑦 2 𝑒 −𝑦 𝑑𝑦
𝑑
⟹ 𝑦 3 𝑒 −𝑦 𝑥 = −2𝑦 2 𝑒 −𝑦 𝑑𝑦
𝑑𝑦
⟹ 𝑦 𝑒 𝑥 = −2 𝑦 2 𝑒 −𝑦 𝑑𝑦
3 −𝑦
⟹ 𝑦 3 𝑒 −𝑦 𝑥 = −2 −𝑦 2 𝑒 −𝑦 − −2𝑦𝑒 −𝑦 𝑑𝑦
⟹ 𝑦 3 𝑒 −𝑦 𝑥 = −2 −𝑦 2 𝑒 −𝑦 + 2 𝑦𝑒 −𝑦 𝑑𝑦
⟹ 𝑦 3 𝑒 −𝑦 𝑥 = −2 −𝑦 2 𝑒 −𝑦 + 2 −𝑦𝑒 −𝑦 − −𝑒 −𝑦 𝑑𝑦
⟹ 𝑦 3 𝑒 −𝑦 𝑥 = −2 −𝑦 2 𝑒 −𝑦 + 2 −𝑦𝑒 −𝑦 − 𝑒 −𝑦
⟹ 𝑦 3 𝑒 −𝑦 𝑥 = −2 −𝑦 2 𝑒 −𝑦 − 2𝑦𝑒 −𝑦 − 2𝑒 −𝑦 + 𝑐
⟹ 𝑦 3 𝑒 −𝑦 𝑥 = 2𝑦 2 𝑒 −𝑦 + 4𝑦𝑒 −𝑦 + 4𝑒 −𝑦 + 𝑐
Thus a family of solutions is defined implicitly by
𝑥𝑦 3 = 2𝑦 2 + 4𝑦 + 4 + 𝑐𝑒 𝑦 ∎
3. 𝑦 + 1 𝑑𝑥 + 4𝑥 − 𝑦 𝑑𝑦 = 0
Solution:
This is linear with respect to 𝑥. Thus, the standard form of our
𝑑𝑥 4 𝑦
equation is + 𝑥= 1
𝑑𝑦 𝑦+1 𝑦+1
Now, solve the integrating factor;
4
𝑑𝑦 4
𝑒 𝑃 𝑑𝑦 = 𝑒 = 𝑒 4 ln 𝑦+1 = 𝑒 ln 𝑦+1 = 𝑦 + 1
𝑦+1 4
Then, multiply the integrating factor to 1 ;
4 𝑑𝑥 4 𝑦
𝑦 +1 + 𝑥=
𝑑𝑦 𝑦+1 𝑦+1
4 𝑑𝑥
⟹ 𝑦 +1 + 4 𝑦 + 1 3𝑥 = 𝑦 + 1 3𝑦
𝑑𝑦
𝑑𝑥
⟹ 𝑦 + 1 4 + 4 𝑦 + 1 3𝑥 = 𝑦 + 1 3𝑦
𝑑𝑦
𝑑
⟹ 𝑦 + 1 4𝑥 = 𝑦 + 1 3𝑦 𝑑𝑦
𝑑𝑦
⟹ 𝑦 + 1 4 𝑥 = 𝑦 + 1 3 𝑦 𝑑𝑦
4𝑥 𝑦 𝑦+1 4 𝑦+1 4
⟹ 𝑦+1 = −
4 4
4𝑥 𝑦 𝑦+1 4 𝑦+1 5
⟹ 𝑦+1 = − +𝑐
4 20
4𝑥 𝑦 𝑦+1 4 𝑦+1 5 −4
⟹ 𝑦+1 = − + 𝑐 20 𝑦 + 1
4 20
⟹ 20𝑥 = 5𝑦 − 𝑦 − 1 + 𝑐 𝑦 + 1 −4
⟹ 20𝑥 = 4𝑦 − 1 + 𝑐 𝑦 + 1 −4 ∎
4. 𝑦 ′ − 𝑦 tan 𝑥 = sin 𝑥 when 𝑦 0 = 1
Solution :
𝑑𝑦
The equation can be also written as − 𝑦 tan 𝑥 = sin 𝑥 1 .
𝑑𝑥
Observe that 1 is linear with respect to 𝑦 and 1 is written already
in standard form. Hence, the integrating factor is;
𝑒 𝑃 𝑑𝑥 = 𝑒 − tan 𝑥 𝑑𝑥
= 𝑒 − tan 𝑥 𝑑𝑥
sin 𝑥
= 𝑒 − cos 𝑥 𝑑𝑥
= 𝑒 − − ln cos 𝑥
= cos 𝑥
Now, multiply the integrating factor to 1 and we get
𝑑𝑦
cos 𝑥 − 𝑦 tan 𝑥 = sin 𝑥
𝑑𝑥
⟹ cos 𝑥 𝑑𝑦 − 𝑦𝑠𝑖𝑛 𝑥 𝑑𝑥 = sin 𝑥 𝑑𝑥
𝑑
⟹ 𝑐𝑜𝑠 𝑥 𝑦 = 𝑠𝑖𝑛 𝑥 𝑑𝑥
𝑑𝑥
𝑑
⟹ 𝑐𝑜𝑠 𝑥 𝑦 = 𝑠𝑖𝑛 𝑥 𝑑𝑥
𝑑𝑥
⟹ cos 𝑥 𝑦 = cos 𝑥 sin 𝑥 𝑑𝑥
𝑠𝑖𝑛2 𝑥
⟹ cos 𝑥 𝑦 = +𝑐
2
2
⟹ 2 cos 𝑥 𝑦 = 𝑠𝑖𝑛 𝑥 + 𝑐
Since we are solving for the particular solution by using the initial value
𝑦 0 = 1. Then, we have
2 cos 𝑥 𝑦 = 𝑠𝑖𝑛2 𝑥 + 𝑐
⟹ 2(1) 1 = 𝑐
⟹ 2=𝑐
Therefore, the particular solution is2 cos 𝑥 𝑦 = 𝑠𝑖𝑛2 𝑥 + 2∎
