Question

Solve the differential equation dy dx + 12x2y = 36x2. SOLUTION The given equation is linear since it has the form of this equation with P(x) = 12x2 and Q(x) = 36x2. An integration factor is I(x) = e∫12x2 dx = .

Answer 1

Answer:

Therefore the solution of the differential equation is

$ye^{4x^3} = 3e^{4x^3}+c$    [ where c is arbitrary constant]

Step-by-step explanation:

Given differential equation is

$\frac{dy}{dx} +12x^2y= 36x^2$

Here $P(x)= 12x^2$   and  $Q(x) = 36 x^2$

The integrating factor of the differential equation is

$= e^{\int P(x) dx$

$=e^{\int 12x^2dx$

$=e^{ \frac{12x^3}{3}}$

$=e^{4x^3}$

Multiplying the integrating factor both sides of the  differential equation

$e^{4x^3}\frac{dy}{dx} +12x^2ye^{4x^3}= 36x^2e^{4x^3}$

$\Rightarrow e^{4x^3} dy+12x^2ye^{4x^3}dx= 36x^2e^{4x^3}dx$

Integrating both sides,

$\int e^{4x^3} dy+\int12x^2ye^{4x^3}dx= \int36x^2e^{4x^3}dx$......(1)

Let

$I= \int36x^2e^{4x^3}dx$

  $= \int3. 12 x^2e^{4x^3}dx$

putting ${4x^3}=z$ , $12x^2 dx=dz$

  $=\int 3. e^zdz$

  $=3e^z+c$     [ where c is arbitrary constant]

Putting the value of z

  $=3e^{4x^3}+c$

From (1) we get

$\int e^{4x^3} dy+\int12x^2ye^{4x^3}dx= \int36x^2e^{4x^3}dx$

$\Rightarrow ye^{4x^3} = 3e^{4x^3}+c$

Therefore the solution of the differential equation is

$ye^{4x^3} = 3e^{4x^3}+c$    [ where c is arbitrary constant]