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UP TGT Mathematics 2019 Official Paper

Option 1 : 0

\(\frac{{dy}}{{dx}} + 4xy = {x^3}\)

L.D.E

\(\frac{{dy}}{{dx}} + Py = Q\)

P = 4x, Q = x^{3}

\(If = {e^{\smallint Pdx}} = {e^{\smallint 4x\;dx}} = {e^{4\smallint x\;dx}} = {e^{4\frac{{{x^2}}}{2}}} = {e^{2{x^2}}}\)

Solution is

\(y{e^{2{x^2}}} = \smallint {x^3}.{e^{2{x^2}}}dx + C\)

Let x^{2} = t ⇒ 2x dx = dt

\( = \smallint {x^2}.x.{e^{2{x^2}}}dx + C\)

\( = \smallint t.{e^{2t}}\frac{{dt}}{2} + C\)

\( = \frac{1}{2}\smallint t.{e^{2t}} + C\)

\( = \frac{1}{2}[\smallint {e^{2t}}\)

\( = \frac{1}{2}\left[ {t.\frac{{{e^{2t}}}}{2} - \smallint 1.\frac{{{e^{2t}}}}{2}dt} \right] + C\)

\(\frac{{t.{e^{2t}}}}{4} - \frac{1}{4}.\frac{{{e^{2t}}}}{2} + C\)

\( = \frac{{t{e^{2t}}}}{4} - \frac{1}{8}\;{e^{2t}} + C\)

Put t = x^{2}

\(y.{e^{2{x^2}}} = \frac{{{x^2}{e^{2x}}}}{4} - \frac{1}{8}{e^{2{x^2}}} + C\)

Divide the whole equation by e^{2x}

\(y = \frac{{{x^2}}}{4} - \frac{1}{8} + C{e^{ - 2{x^2}}}\)

Given y(0) = 0

\(0 = 0 - \frac{1}{8} + C.{e^{ - 0}}\)

\(C = \frac{1}{8}\)

\(y.{e^{2{x^2}}} = \frac{{{x^2}{e^{2x}}}}{4} - \frac{1}{8}{e^{2{x^2}}} + \frac{1}{8}\)

\(y = \frac{{{x^2}}}{4} - \frac{1}{8} + \frac{1}{8}{e^{ - 2{x^2}}}\)

\(\mathop {\lim }\limits_{x \to 0} 0 - \frac{1}{8} + \frac{1}{8}\left( 1 \right)\)

\(\mathop {\lim }\limits_{x \to 0} y = 0\)

CT 1: हिन्दी (आदिकाल)

6388

10 Questions
40 Marks
10 Mins