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

Therefore the solution of the differential equation is

`ye^(4x^3) = 3e^(4x^3)+c` [ where c is arbitrary constant]

Explanation:

Given differential equation is

`(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^{ (12x^3)/(3)}`

`=e^(4x^3)`

Multiplying the integrating factor both sides of the differential equation

`e^(4x^3)(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]