If the solution curve of the differential equation (2x - 10y3)dy + ydx = 0, passes through the points (0, 1) and (2, \beta ), then \beta is a root of the equation :

Differential Equations PYQ — JEE Main 2021 · 27 Aug · Shift 2 · Q27 with Solution

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Correct answer: D

Write the differential equation clearly

Given:

(2x-10y^3)\,dy + y\,dx = 0

Rearrange as

y\,dx + (2x-10y^3)\,dy=0

This is of the form

M(x,y)\,dx + N(x,y)\,dy=0

with

M=y, \qquad N=2x-10y^3

Check if the equation is exact

For exactness, we need

\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}

Now,

\frac{\partial M}{\partial y}=1, \qquad \frac{\partial N}{\partial x}=2

So it is not exact.

Treat $x$ as a function of $y$

From

(2x-10y^3)\,dy + y\,dx = 0

divide by $dy$ :

y\frac{dx}{dy} + 2x - 10y^3 = 0

So,

\frac{dx}{dy} + \frac{2}{y}x = 10y^2

This is a linear differential equation in $x$ as a function of $y$ .

Find the integrating factor

IF = e^{\int \frac{2}{y}dy} = e^{2\ln y}=y^2

Multiply the equation by $y^2$ :

y^2\frac{dx}{dy} + 2yx = 10y^4

The left side is

\frac{d}{dy}(xy^2)

Hence,

\frac{d}{dy}(xy^2)=10y^4

Integrate

xy^2 = \int 10y^4\,dy = 2y^5 + C

So the solution curve is

xy^2 = 2y^5 + C

xy^2 - 2y^5 = C

Use the point $(0,1)$

Substitute $x=0, y=1$ :

0\cdot 1^2 - 2\cdot 1^5 = C

C=-2

Thus the particular solution is

xy^2 - 2y^5 = -2

Use the point $(2,\beta)$

Substitute $x=2, y=\beta$ :

2\beta^2 - 2\beta^5 = -2

Divide by $2$ :

\beta^2 - \beta^5 = -1

Rearrange:

\beta^5 - \beta^2 - 1 = 0

So $\beta$ is a root of

y^5 - y^2 - 1 = 0

Match with options

This corresponds to:

Option D:

y^5 - y^2 - 1 = 0

Therefore, the correct answer is D.

More from Differential Equations

View written solutionFree

Correct answer: D

Write the differential equation clearly

Given:

(2x-10y^3)\,dy + y\,dx = 0

Rearrange as

y\,dx + (2x-10y^3)\,dy=0

This is of the form

M(x,y)\,dx + N(x,y)\,dy=0

with

M=y, \qquad N=2x-10y^3

Check if the equation is exact

For exactness, we need

\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}

Now,

\frac{\partial M}{\partial y}=1, \qquad \frac{\partial N}{\partial x}=2

So it is not exact.

Treat $x$ as a function of $y$

From

(2x-10y^3)\,dy + y\,dx = 0

divide by $dy$ :

y\frac{dx}{dy} + 2x - 10y^3 = 0

So,

\frac{dx}{dy} + \frac{2}{y}x = 10y^2

This is a linear differential equation in $x$ as a function of $y$ .

Find the integrating factor

IF = e^{\int \frac{2}{y}dy} = e^{2\ln y}=y^2

Multiply the equation by $y^2$ :

y^2\frac{dx}{dy} + 2yx = 10y^4

The left side is

\frac{d}{dy}(xy^2)

Hence,

\frac{d}{dy}(xy^2)=10y^4

Integrate

xy^2 = \int 10y^4\,dy = 2y^5 + C

So the solution curve is

xy^2 = 2y^5 + C

xy^2 - 2y^5 = C

Use the point $(0,1)$

Substitute $x=0, y=1$ :

0\cdot 1^2 - 2\cdot 1^5 = C

C=-2

Thus the particular solution is

xy^2 - 2y^5 = -2

Use the point $(2,\beta)$

Substitute $x=2, y=\beta$ :

2\beta^2 - 2\beta^5 = -2

Divide by $2$ :

\beta^2 - \beta^5 = -1

Rearrange:

\beta^5 - \beta^2 - 1 = 0

So $\beta$ is a root of

y^5 - y^2 - 1 = 0

Match with options

This corresponds to:

Option D:

y^5 - y^2 - 1 = 0

Therefore, the correct answer is D.

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