QJEEM21Mab8e3: JEE Main 2021 Mathematics - Class 12 JEEM21Mab8e3 Solution: If the curve y = y(x) is the solution of the differential…

Question

Verified Solution • 2025–26 Session moderate

If the curve y = y(x) is the solution of the differential equation

2({x^2} + {x^{5/4}})dy - y(x + {x^{1/4}})dx = {2x^{9/4}}dx

, x > 0 which

passes through the point

\left( {1,1 - {4 \over 3}{{\log }_e}2} \right)

, then the value of y(16) is equal to :

Accepted Answer

{{dy} \over {dx}} - {y \over {2x}} = {{{x^{9/4}}} \over {{x^{5/4}}({x^{3/4}} + 1)}}

IF = {e^{ - \int {{{dx} \over {2x}}} }} = {e^{ - {1 \over 2}\ln x}} = {1 \over {{x^{1/2}}}}

y.{x^{ - 1/2}} = \int {{{{x^{9/4}}.{x^{ - 1/2}}} \over {{x^{5/4}}({x^{3/4}} + 1)}}} dx

=

\int {{{{x^{1/2}}} \over {({x^{3/4}} + 1)}}} dx

Let,

x = {t^4} \Rightarrow dx = 4{t^3}dt

=

\int {{{{t^2}.4{t^3}dt} \over {({t^3} + 1)}}}

=

4\int {{{{t^2}({t^3} + 1 - 1)} \over {({t^3} + 1)}}} d

=

4\int {{t^2}dt - 4\int {{{{t^2}} \over {{t^3} + 1}}dt} }

=

{{4{t^3}} \over 3} - {4 \over 3}\ln ({t^3} + 1) + C

y{x^{ - 1/2}} = {{4{x^{3/4}}} \over 3} - {4 \over 3}\ln ({x^{3/4}} + 1) + C

It passes through the point

\left( {1,1 - {4 \over 3}{{\log }_e}2} \right)

\therefore

1 - {4 \over 3}{\log _e}2 = {4 \over 3} - {4 \over 3}{\log _e}2 + C

\Rightarrow C = - {1 \over 3}

y = {4 \over 3}{x^{5/4}} - {4 \over 3}\sqrt x \ln ({x^{3/4}} + 1) - {{\sqrt x } \over 3}

y(16) = {4 \over 3} \times 32 - {4 \over 3} \times 4\ln 9 - {4 \over 3}

= {{124} \over 3} - {{32} \over 3}\ln 3 = 4\left( {{{31} \over 3} - {8 \over 3}\ln 3} \right)

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