Q1. i) The above equations can be written as That is Solve by substitution Therefore, That is Integrate both sides ii) Solve the integral on the LHS by trigonometric...

Q1.

$xy'=y+\sqrt{x^2+y^2}$

The above equations can be written as

$xy'-y=\sqrt{x^2+y^2}$

That is

$y'-\frac{y}{x}=\frac{\sqrt{x^2+y^2}}{x}=\sqrt{1+(\frac{y}{x})^2}$

Solve by substitution

$Let, y=ux => y'=(ux)'=u'x+u(x)'=u'x+u$

Therefore,

$u'x+u-u=\sqrt{1+u^2} => u'x=\sqrt{1+u^2}$

That is

$\frac{du}{\sqrt{1+u^2}}=\frac{dx}{x}$

Integrate both sides

$\int \frac{du}{\sqrt{1+u^2}}=\int \frac{dx}{x}=lnx+C$

ii)

Solve the integral on the LHS by trigonometric substitution

$Let, u=tan\theta => du=sec^2\theta d\theta$

Therefore,

$\int \frac{sec^2\theta }{\sqrt{1+tan^2\theta}}d\theta=\int sec\theta d\theta=ln|sec\theta+tan\theta|$

Substitute sec(theta) and tan(theta) in terms of u

$ln|u+\sqrt{1+u^2}|$

Substitute u in terms of y and x implies

$ln|\frac{y}{x}+\sqrt{1+(\frac{y}{x})^2}|$

iii)

Therefore, the solution becomes

$ln|\frac{y}{x}+\sqrt{1+(\frac{y}{x})^2}|=lnx+C => ln|y+\sqrt{x^2+y^2}|-lnx=lnx+C$

Simplify

$ln|y+\sqrt{x^2+y^2}|=lnx^2+C => ln(\frac{y+\sqrt{x^2+y^2}}{x^2})=C$

That is

$\frac{y+\sqrt{x^2+y^2}}{x^2}=e^C=K => y+\sqrt{x^2+y^2}=Kx^2$

iv)

Simplify

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

Squaring both sides

$\left ( \sqrt{x^2+y^2} \right )^2=\left ( kx^2-y \right )^2 => x^2+y^2=k^2x^4+y^2-2kx^2y$

$x^2=k^2x^4-2kx^2y => 2kx^2y=k^2x^4-x^2 => 2ky=k^2x^2-1$

Hence, the solution is

$y=\frac{k^2x^2-1}{2k}$

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