Question

dy/dx+ycosx=y^3sin2x​

Answer 1

Step-by-step explanation:

give dy/dx+ycosx=y^3sin2x-------'1'

Above equation is

1/y^3 dy/dx +cosx (1/y^2)=sin 2x-------''2''

1/y^2=Z

-2/y^3 dy/dx=dz/dx

1/y^3 dy/dx = -1/2 dz/dx

Substituting the value in2 we get

-1/2 dz/dx +cosx.z=sin2x

z. dz/dx -2cosx.z=-2sin2x----------"3"

"3" is of the form dz/dx+p(x) z=Q(x)

p(x) =-2cosx, Q(x) = -2sin2x

I. F =e^-2sinx

where sinx=t

cosxdx= dt

=e^t(1-t)+c

1/y^2 e^-2sinx =(1+2sinx)e^-2sinx+c

Answer 2

The solution of the differential equation is $\bf \frac{1}{ {y}^{2} } {e}^{ - 2sinx } = (2 \: sin \: x +1) {e}^{ - 2sin \: x} + C$

Given:

• $\frac{dy}{dx} + y \: cos \: x = {y}^{3}sin \: 2x$

To find:

• Find the solution of differential equation.

Solution:

Formula to be used:

• Linear differential equation:$\bf \frac{dy}{dx} + Py = Q...eq1$
• Solution of this differential equation$\boxed{\bf y. \:I.F. = \int \: Q. \: I.F.\: dx}$here $I.F. = {e}^{ \int P \: dx}$
• Integration by parts: $\bf \int UV dx=U\int V dx-\int\left(\frac{dU}{dx}\int\: Vdx\right)dx$

Step 1:

Simplify the equation.

Divide the equation by y³.

$\frac{1}{ {y}^{3} } \frac{dy}{dx} + \frac{1}{ {y}^{2} } \: cos \: x =sin \: 2x$

Let

$\frac{1}{ {y}^{2} } = t$

$- \frac{2}{ {y}^{3} } \frac{dy}{dx} = \frac{dt}{dx}$

Substitute the values in the equation.

$- \frac{1}{2} \frac{dt}{dx} + t \: cosx = sin \: 2x$

or

$\frac{dt}{dx} - 2 t \: cosx = - 2sin \: 2x ...eq2$

Compare the equation 2 with standard equation.

It is clear that

$\bf P = - 2 \: cos \: x$

$\bf Q = - 2 \: sin \: 2x$

Step 2:

Find the solution of linear equation.

Calculate I.F.

$I.F. = {e}^{ \int \: P \: dx}$

$I.F. = {e}^{ \int - 2cosx \: dx}$

$I.F. = {e}^{ - 2sin \: x } ...eq3$

Now the solution of differential equation is

$t \: (I.F.) = \int \:Q.( I.F.) \: dx$

Put the value of IF from eq3 .

$t {e}^{ - 2sinx } = \int \: - 2sin \: 2x \: {e}^{ - 2 \: cos \: x} dx...eq4$

Step 3:

Substitute the RHS to integrate.

Rewrite eq4

As we know that sin2x = 2sinx cos x

$t {e}^{ - 2sinx } = \int \: - 2sin \: x \: .2cos \: x . \: {e}^{ - 2 \: sin \: x} dx$

Let

$- 2 \: sin \: x = v$

$- 2 \: cos \: x \: dx = dv$

Substitute the values.

$t {e}^{ - 2sinx } = - \int \: v\: {e}^{v} dv$

Apply integration by parts in RHS.

$t {e}^{ - 2sinx } = - v \int{e}^{v} dv + \int \: \left( \frac{dv}{dv} \int{e}^{v} dv \right)dv$

$t {e}^{ - 2sinx } = - v {e}^{v} + \int \: 1.{e}^{v} dv$

$t {e}^{ - 2sinx } = - v {e}^{v} + {e}^{v} + C$

Undo substitution

$t {e}^{ - 2sinx } = 2 \: sin \: x {e}^{ - 2 \: sin \: x} + {e}^{ - 2sin \: x} + C$

Put the value of t also.

$\frac{1}{ {y}^{2} } {e}^{ - 2sinx } = 2 \: sin \: x {e}^{ - 2 \: sin \: x} + {e}^{ - 2sin \: x} + C$

$\frac{1}{ {y}^{2} } {e}^{ - 2sinx } = (2 \: sin \: x +1) {e}^{ - 2sin \: x} + C$

Thus,

The solution of the differential equation is $\bf \frac{1}{ {y}^{2} } {e}^{ - 2sinx } = (2 \: sin \: x +1) {e}^{ - 2sin \: x} + C$

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