Let the solution curve of the differential equation x ? d y - y ? d x = x 2 + y 2 ? d x , x 0 , y ( 1 ) = 0 be y = y ( x ) . Then y ( 3 ) is equal to [2026]

Question

Let the solution curve of the differential equation $x d y - y d x = \sqrt{x^{2} + y^{2}} d x, x > 0,$ $y (1) = 0$ be $y = y (x)$ . Then $y (3)$ is equal to [2026]

Smart Achievers · Accepted Answer

(4)

$\frac{x d y - y d x}{x^{2}} = \frac{\sqrt{x^{2} + y^{2}}}{x^{2}} d x$

$d (\frac{y}{x}) = \sqrt{1 + \frac{y^{2}}{x^{2}}} \cdot \frac{1}{x} d x$

$\int \frac{d (\frac{y}{x})}{\sqrt{1 + {(\frac{y}{x})}^{2}}} = \int \frac{1}{x} d x$

$= l n (\frac{y}{x} + \sqrt{1 + \frac{y^{2}}{x^{2}}}) = l n x + l n k = l n k x$

$\Rightarrow y + \sqrt{x^{2} + y^{2}} = k x^{2}$

$0 + 1 = k$

$\Rightarrow y + \sqrt{x^{2} + y^{2}} = x^{2}$

$y + \sqrt{9 + y^{2}} = 9$

$y = 4$

Solution

Q.
Let the solution curve of the differential equation $x d y - y d x = \sqrt{x^{2} + y^{2}} d x, x > 0,$ $y (1) = 0$ be $y = y (x)$ . Then $y (3)$ is equal to [2026]

1 1

2 2

3 6

4 4

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Solution

Q. Let the solution curve of the differential equation x dy-y dx=x2+y2 dx, x>0, y(1)=0 be y=y(x). Then y(3) is equal to [2026]

1 1

2 2

3 6

4 4

Ans. (4) x dy-y dxx2=x2+y2x2dx d(yx)=1+y2x2·1xdx ∫d(yx)1+(yx)2=∫1xdx =ln(yx+1+y2x2)=lnx+lnk=lnkx ⇒y+x2+y2=kx2 0+1=k ⇒y+x2+y2=x2 y+9+y2=9 y=4

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Q.
Let the solution curve of the differential equation $x d y - y d x = \sqrt{x^{2} + y^{2}} d x, x > 0,$ $y (1) = 0$ be $y = y (x)$ . Then $y (3)$ is equal to [2026]