The solution of the differential equation {{{{dy}}}}{{{{dx}}}} = {{{{y}} '≤ft( {{x}} ) - {{{y}}²}}{{ ≤ft( {{x}} )}} is

Correct Answer is : y=ϕ(x)x+c{{y}} = {{ ≤ft( {{x}} )}}{{{{x}} + {{c}}}}y=x+cϕ(x)

{{y}} = {{{x}}}{{ ≤ft( {{x}} ) + {{c}}}}

{{y}} = {{ ≤ft( {{x}} )}}{{{x}}} + {{c}}

{{y}} = {{ ≤ft( {{x}} ) + {{c}}}}{{{x}}}

{{y}} = {{ ≤ft( {{x}} )}}{{{{x}} + {{c}}}}

Correct Answer is : y=ϕ(x)x+c{{y}} = {{ ≤ft( {{x}} )}}{{{{x}} + {{c}}}}y=x+cϕ(x)

Test Index

UPSC NDA 2 2017 Mathematics Solved Paper

Question : 94 of 120

Marks: +1, -0

\frac{{{\rm{dy}}}}{{{\rm{dx}}}} = \frac{{{\rm{y}}\phi '\left( {\rm{x}} \right) - {{\rm{y}}^2}}}{{\phi \left( {\rm{x}} \right)}}

Solution:

Concept:

Use the substitution

z = \frac{y}{\phi(x)}

to separate variables and solve.

Explanation:

Start with

\frac{dy}{dx} = \frac{y\phi'(x) - y^2}{\phi(x)}

Rewrite the right side as

\phi'(x) \cdot \frac{y}{\phi(x)} - \phi(x)\left(\frac{y}{\phi(x)}\right)^2

Let

z = \frac{y}{\phi(x)}

, so

y = \phi(x)z

Differentiate:

\frac{dy}{dx} = \phi(x)\frac{dz}{dx} + \phi'(x)z

Substitute into the equation:

\phi'(x)z - \phi(x)z^2 = \phi(x)\frac{dz}{dx} + \phi'(x)z

Cancel

\phi'(x)z

on both sides:

-\phi(x)z^2 = \phi(x)\frac{dz}{dx}

\phi(x) \neq 0

, divide by

\phi(x)

-z^2 = \frac{dz}{dx}

Separate variables:

-\frac{dz}{z^2} = dx

Integrate:

\int -\frac{dz}{z^2} = \int dx

gives

\frac{1}{z} = x + c

Replace

z

\frac{\phi(x)}{y} = x + c

Solve for

y

y = \frac{\phi(x)}{x + c}

Answer:

y = \frac{\phi(x)}{x + c}

(option D).

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