Question

dy/dx + 2xy =x^3 solve the differential equation

Answer 1

Answer :

Given differential equation is

$\frac{dy}{dx} + 2xy = {x}^{3}$ ...(i)

Then, Integrating Factor

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

= $e^{x^{2}}$

Now, multiplying (i) by I.F., we get

$\frac{d}{dx} (y {e}^{ {x}^{2} } ) = {x}^{3} {e}^{ {x}^{2} }$

Taking integration, we get

$y {e}^{{x}^{2}} = \int {x}^{3} {e}^{ {x}^{2} } dx$ ...(ii)

Let, $x^{2}$ = z

$\implies$ 2x dx = dz

Thus,

$\int {x}^{3} {e}^{ {x}^{2} }dx$

= $\frac{1}{2} \int {x}^{2} {e}^{x^{2}} (2xdx)$

= $\frac{1}{2} \int z e^{z} dz$

$= \frac{1}{2} z \int {e}^{z} dz - \frac{1}{2} \int ( \frac{d}{dz} (z) \times \int {e}^{z} dz)dz \\ \\ = \frac{1}{2} z {e}^{z} - \frac{1}{2} \int {e}^{z} dz + c$

where c is integral constant

$= \frac{1}{2} z {e}^{z} - \frac{1}{2} {e}^{z} + c \\ \\ = \frac{1}{2} {e}^{z} (z - 1) + c \\ \\ = \frac{1}{2} {e}^{ {x}^{2} } ( {x}^{2} - 1) + c$

From (ii), we get

$y {e}^{ {x}^{2} } = \frac{1}{2} {e}^{ {x}^{2} } ( {x}^{2} - 1) + c \\ \\ \implies y = \frac{1}{2} ( {x}^{2} - 1) + c \: {e}^{ - {x}^{2} }$

which is the required solution

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