1. Rewrite the differential equation

Given

$$(x^2+y^2)\,dx-5xy\,dy=0$$

with initial condition $y(1)=0$.

Divide by $dx$:

$$(x^2+y^2)-5xy\frac{dy}{dx}=0$$

so

$$\frac{dy}{dx}=\frac{x^2+y^2}{5xy}.$$

This is a homogeneous differential equation.

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2. Use the substitution

Let

$$y=vx \quad \Rightarrow \quad \frac{dy}{dx}=v+x\frac{dv}{dx}.$$

Also,

$$y^2=v^2x^2.$$

Substitute into the DE:

$$v+x\frac{dv}{dx}=\frac{x^2+v^2x^2}{5x(vx)}=\frac{1+v^2}{5v}.$$

Hence,

$$x\frac{dv}{dx}=\frac{1+v^2}{5v}-v$$ $$=\frac{1+v^2-5v^2}{5v}=\frac{1-4v^2}{5v}.$$

So,

$$\frac{5v}{1-4v^2}\,dv=\frac{dx}{x}.$$

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3. Integrate

Integrate both sides:

$$\int \frac{5v}{1-4v^2}\,dv = \int \frac{dx}{x}.$$

Let

$$u=1-4v^2 \quad \Rightarrow \quad d\nu=-8v\,dv.$$

Then

$$\int \frac{5v}{1-4v^2}\,dv n= -\frac{5}{8}\int \frac{d\nu}{\nu} = -\frac{5}{8}\ln|\nu|.$$

Therefore,

$$-\frac{5}{8}\ln|1-4v^2|=\ln|x|+C.$$

Multiply by $-\frac{8}{5}$:

$$\ln|1-4v^2|=-\frac{8}{5}\ln|x|+C_1.$$

Exponentiating,

$$|1-4v^2| = Cx^{-8/5}.$$

Now substitute $v=\frac{y}{x}$:

$$\left|1-4\frac{y^2}{x^2}\right| = Cx^{-8/5}.$$

Multiply by $x^2$ inside the absolute value:

$$\left|\frac{x^2-4y^2}{x^2}\right| = Cx^{-8/5}$$ $$\frac{|x^2-4y^2|}{x^2}=Cx^{-8/5}.$$

Hence,

$$|x^2-4y^2| = Cx^{2/5}.$$

Raise both sides to power $5$:

$$|x^2-4y^2|^5 = C^5x^2.$$

Let $C^5=C_2$:

$$|x^2-4y^2|^5 = C_2x^2.$$

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4. Apply the initial condition

Given $y(1)=0$:

$$|1^2-4\cdot 0^2|^5 = C_2\cdot 1^2$$ $$1=C_2.$$

So the required solution is

$$|x^2-4y^2|^5=x^2.$$

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5. Check options

• A: $|x^2-4y^2|^5=x^2$ ✅
• B: $|x^2-2y^2|^6=x$ ❌
• C: $|x^2-2y^2|^5=x^2$ ❌
• D: $|x^2-4y^2|^6=x$ ❌

Therefore, the correct option is A.