2 methods for solving first order differential equations. One method requires th

Question

2 methods for solving first order differential equations. One method requires that the equation be seperable. The other requires that the equation be linear. Below you will find two first order differential equations. Describe how you will determine which method to use for each equation and then solve it using that method. 1. (2y)e^(-2x) (dy)/(dx)=3 . I will solve this using the method for ____________________________because _______________________________ 2. y'+4xy-x=0 I will solve this using the method for ____________________________because _______________________________. If answered correctly I will rate. thanks ahead of time.

Explanation / Answer

1). This can be done using separation of variables. As can be seen that x and y terms can be separated and taken on two different sides

2ye-2x (dy/dx)= 3

2y.dy= 3e2x.dx

Integrating both the sides, we have

y2= (3e2x/2) + C

2). Here x and y terms cant be separated. So we use the second method.

(dy/dx) + 4xy = x

We will multiply both the sides by the integrating factor. Here the integrrating factor is e^(integration of 4x w.r.t x)

So the integrating factor is e2x^2

So now we have,

e2x^2 (dy/dx) + e2x^2.4xy = e2x^2.x

We see that L.H.S of the equation is derivative of e2x^2.y with respect to x

So taking the integration of both the sides with respect to x, we get

e2x^2.y= Integration ( e2x^2.x) dx + C

[You can solve the integration on the R.H.S by substitution method i.e by taking x2= t, so 2xdx=dt and so we weill have

int (e2x^2.x)dx = int (e2t)dt/2

= e2t/4

= e2x^2/4]

SO e2x^2.y = (e2x^2/4) + C

==> y = (1/4) + C.e-2x^2

Related Questions

Navigate

• Browse (All)
• Browse 2
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.