Let y = y(x) be the solution curve of the differential equation <math><mrow><mi>s</mi><mi>i</mi><mi>n</mi><mrow><mo>(</mo><mrow><mn>2</mn><msup><mrow><mi>x</mi></mrow><mrow><mn>2</mn></mrow></msup></mrow><mo>)</mo></mrow><mi>l</mi><mi>o</mi><msub><mrow><mi>g</mi></mrow><mrow><mi>e</mi></mrow></msub><mrow><mo>(</mo><mrow><mi>t</mi><mi>a</mi><mi>n</mi><msup><mrow><mi>x</mi></mrow><mrow><mn>2</mn></mrow></msup></mrow><mo>)</mo></mrow><mi>d</mi><mi>y</mi><mo>+</mo><mrow><mo>(</mo><mrow><mn>4</mn><mi>x</mi><mi>y</mi><mo>−</mo><mn>4</mn><mroot><mrow><mn>2</mn></mrow><mrow></mrow></mroot><mi>x</mi><mi>s</mi><mi>i</mi><mi>n</mi><mrow><mo>(</mo><mrow><msup><mrow><mi>x</mi></mrow><mrow><mn>2</mn></mrow></msup><mo>−</mo><mfrac><mrow><mi>π</mi></mrow><mrow><mn>4</mn></mrow></mfrac></mrow><mo>)</mo></mrow></mrow><mo>)</mo></mrow><mi>d</mi><mi>x</mi><mo>=</mo><mn>0</mn><mo>,</mo></mrow></math> <math><mrow><mn>0</mn><mo><</mo><mi>x</mi><mo><</mo><mroot><mrow><mfrac><mrow><mi>π</mi></mrow><mrow><mn>2</mn></mrow></mfrac><mo>,</mo></mrow><mrow></mrow></mroot></mrow></math> which passes through the point <math><mrow><mrow><mo>(</mo><mrow><mroot><mrow><mfrac><mrow><mi>π</mi></mrow><mrow><mn>6</mn></mrow></mfrac><mo>,</mo><mtext> </mtext><mtext> </mtext><mn>1</mn></mrow><mrow></mrow></mroot></mrow><mo>)</mo></mrow></mrow></math> . Then <math><mrow><mrow><mo>|</mo><mrow><mi>y</mi><mrow><mo>(</mo><mrow><mroot><mrow><mfrac><mrow><mi>π</mi></mrow><mrow><mn>3</mn></mrow></mfrac></mrow><mrow></mrow></mroot></mrow><mo>)</mo></mrow></mrow><mo>|</mo></mrow></mrow></math> is equal to

sin(2x2).109e(tanx2)dy+4xy dx=42.x.(sinx2cosπ4−cosx2sinπ4)dx⇒ln(tanx2)dy+4xsin(2x2)ydx=4x(sinx2−cosx2)sin(2x2)dxIntegrate⇒y.ln(tanx2)=2.ln(sinx2+cosx2−1sinx2+cosx2+1)+Cx=π6,y=1 calculate C.

Ask & Answer Question

First Order Differential Equation Differential Equations Maths Class 12th

Let y = y(x) be the solution curve of the differential equation $s i n (2 x^{2}) l o g_{e} (t a n x^{2}) d y + (4 x y - 4 \sqrt[]{2} x s i n (x^{2} - \frac{π}{4})) d x = 0,$ $0 < x < \sqrt[]{\frac{π}{2},}$ which passes through the point $(\sqrt[]{\frac{π}{6}, 1})$ . Then $| y (\sqrt[]{\frac{π}{3}}) |$ is equal to

9 Views | Posted 2 months ago

Asked by Shiksha User

1 Answer

Answered by

Payal Gupta | Contributor-Level 10

2 months ago

$s i n (2 x^{2}) . 10 9_{e} (t a n x^{2}) d y + 4 x y d x = 4 \sqrt[]{2} . x . (s i n x^{2} c o s \frac{π}{4} - c o s x^{2} s i n \frac{π}{4}) d x$

$\Rightarrow l n (t a n x^{2}) d y + \frac{4 x}{s i n (2 x^{2})} y d x = \frac{4 x (s i n x^{2} - c o s x^{2})}{s i n (2 x^{2})} d x$

Integrate

$\Rightarrow y . l n (t a n x^{2}) = 2 . l n (\frac{s i n x^{2} + c o s x^{2} - 1}{s i n x^{2} + c o s x^{2} + 1}) + C$

$x = \sqrt[]{\frac{π}{6}}, y = 1$ calculate C.

0 Upvotes 0 Downvotes

1 Answer

Similar Questions for you

Learn more about...

Most viewed information

Share Your College Life Experience

Didn't find the answer you were looking for?

Need guidance on career and education? Ask our experts

Your Question

Let y = y(x) be the solution curve of the differential equation sin(2x2)loge(tanx2)dy+(4xy−42xsin(x2−π4))dx=0, 0<x<π2, which passes through the point (π6, 1) . Then |y(π3)| is equal to