Question

dy/dx+y=3e^xy^3plz solve it​

Answer 1

Step-by-step explanation:

We have,

$\frac{dy}{dx} + y = 3 {e}^{x} {y}^{3}$

$\implies( {y}^{ - 3} )\frac{dy}{dx} + y^{ - 2} = 3 {e}^{x}$

$Let \: \: {y}^{ - 2} = v \\ \implies - 2 {y}^{ - 3} \frac{dy}{dx} = \frac{dv}{dx} \\ \implies {y}^{ - 3} \frac{dy}{dx} = - \frac{1}{2} \frac{dv}{dx}$

$\implies - \frac{1}{2} \frac{dv}{dx} + v = 3 {e}^{x}$

$\implies \frac{dv}{dx} - 2 v = - 6{e}^{x}$

$I.F. = {e}^{ \int( - 2)dx} = {e}^{ - 2x}$

Now,

$v. {e}^{ - 2x} = \int( - 6) {e}^{x} . {e}^{ - 2x} dx$

$\implies \: v. {e}^{ - 2x} = - 6 \int {e}^{ -x} dx$

$\implies \: v. {e}^{ - 2x} = 6 {e}^{ -x} +C$

$\implies \: {y}^{ - 2} = 6 {e}^{ x} +C {e}^{2x}$

Answer 2

Answer:

The solution of the equation is

$y = \frac{1}{ \sqrt{6 {e}^{x} + ce {}^{2x} } }$

Step-by-step explanation:

The given differential equation is

$\frac{dy}{dx} + y = 3 {e}^{x} {y}^{3}$

We shall convert this into a linear differential equation by dividing both sides with the highest degree variable.Hence,

$\frac{1}{ {y}^{3} } . \frac{dy}{dx} + \frac{1}{ {y}^{2} } = 3 {e}^{x}$

Let us assume that,

$\frac{1}{ {y}^{2} } = t \\ = > \frac{ - 2}{ {y}^{3} } \frac{dy}{dx} = \frac{dt}{dx} \\ = > \frac{ 1}{ {y}^{3} } \frac{dy}{dx} = \frac{ - 1}{2} . \frac{dt}{dx}$

Substituting these terms in our differential equation,we get

$- \frac{1}{2} \frac{dt}{dx} + t = 3 {e}^{x}$

Multiplying both the sides with -2,we get

$\frac{dt}{dx} - 2t = - 6 {e}^{x}$

This is an linear differential equation in variable t.

Its integrating factor is

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

Hence the solution of our equation is

$t(I.F)=\int(-6e^{x}(I.F))dx$

Substituting the value of integrating factor in above relation,we get

$t. {e}^{ - 2x} = \int - 6 {e}^{x} ( {e}^{ - 2x} )dx \\ = > t {e}^{ - 2x} = 6 {e}^{ - x} + c \\ = > t = 6 {e}^{x} + ce {}^{2x} \\ = > \frac{1}{ {y}^{2} } = 6 {e}^{x} + ce {}^{2x} \\ = > y = \frac{1}{ \sqrt{6 {e}^{x} + ce {}^{2x} } }$

Therefore,the solution of the equation is

$y = \frac{1}{ \sqrt{6 {e}^{x} + ce {}^{2x} } }$

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