a) Solve: 2x^2\,dy = (x^2+y^2)\,dx . [4] OR (x^2+1) dy

a) Solution of the Differential Equation: $2x^{2}dy=(x^2+y^2)dx$

Step 1: Rewrite the Equation in Standard Form

The given differential equation is: $$2x^{2}dy=(x^2+y^2)dx$$

Divide both sides by $dx$ to express it in the form $\frac{dy}{dx}=f(x,y)$: $$2x^2\frac{dy}{dx}=x^2+y^2$$

Divide both sides by $2x^2$: $$\frac{dy}{dx}=\frac{x^2+y^2}{2x^2}$$

Simplify the right-hand side: $$\frac{dy}{dx}=\frac{1}{2}+\frac{y^2}{2x^2}$$

Step 2: Recognize the Type of Differential Equation

The equation is: $$\frac{dy}{dx}=\frac{1}{2}+\frac{y^2}{2x^2}$$

This resembles a Bernoulli equation of the form: $$\frac{dy}{dx}+P(x)y=Q(x)y^n$$

Rewrite the given equation to match the Bernoulli form: $$\frac{dy}{dx}-\frac{y^2}{2x^2}=\frac{1}{2}$$

Here, $P(x)=-\frac{1}{2x^2}$, $Q(x)=\frac{1}{2}$, and $n=2$.

Step 3: Substitution to Solve the Bernoulli Equation

For a Bernoulli equation, use the substitution: $$v=y^{1-n}$$ Since $n=2$, we have: $$v=y^{1-2}=y^{-1}⟹v=\frac{1}{y}$$

Differentiate $v$ with respect to $x$: $$\frac{dv}{dx}=-\frac{1}{y^2}\frac{dy}{dx}⟹\frac{dy}{dx}=-y^2\frac{dv}{dx}$$

Substitute $\frac{dy}{dx}$ and $y=\frac{1}{v}$ into the original equation: $$-y^2\frac{dv}{dx}-\frac{y^2}{2x^2}=\frac{1}{2}$$

Substitute $y=\frac{1}{v}$: $$-\frac{1}{v^2}\frac{dv}{dx}-\frac{1}{2x^2{v}^2}=\frac{1}{2}$$

Multiply through by $-v^2$ to simplify: $$\frac{dv}{dx}+\frac{1}{2x^2}=-\frac{v^2}{2}$$

Rearrange: $$\frac{dv}{dx}+\frac{v^2}{2}=-\frac{1}{2x^2}$$

Step 4: Solve the Resulting First-Order Linear Equation

The equation is now: $$\frac{dv}{dx}+\frac{v^2}{2}=-\frac{1}{2x^2}$$

This is a Riccati equation, but it can also be approached by recognizing it as a separable equation after rearrangement. However, let's use an integrating factor approach.

Rewrite the equation: $$\frac{dv}{dx}+\frac{v^2}{2}=-\frac{1}{2x^2}$$

This resembles the form: $$\frac{dv}{dx}+P(x)v=Q(x)$$

But here, the term $\frac{v^2}{2}$ complicates things. Instead, let's consider an alternative substitution or recognize it as a special case.

Step 5: Alternative Approach - Homogeneous Equation

Let's check if the equation is homogeneous. Rewrite the original equation: $$2x^{2}dy=(x^2+y^2)dx$$

Divide both sides by $x^2$: $$2\frac{dy}{dx}=1+{\left(\frac{y}{x}\right)}^2$$

Let $v=\frac{y}{x}$, so $y=vx$ and $\frac{dy}{dx}=v+x\frac{dv}{dx}$.

Substitute into the equation: $$2\left(v+x\frac{dv}{dx}\right)=1+v^2$$

Simplify: $$2v+2x\frac{dv}{dx}=1+v^2$$

Rearrange to isolate $\frac{dv}{dx}$: $$2x\frac{dv}{dx}=1+v^2-2v$$ $$\frac{dv}{dx}=\frac{1+v^2-2v}{2x}$$

Separate variables: $$\frac{dv}{1+v^2-2v}=\frac{dx}{2x}$$

Step 6: Integrate Both Sides

First, simplify the left-hand side: $$1+v^2-2v=v^2-2v+1=(v-1{)}^2$$

So the equation becomes: $$\frac{dv}{(v-1{)}^2}=\frac{dx}{2x}$$

Integrate both sides: $$\int \frac{dv}{(v-1{)}^2}=\int \frac{dx}{2x}$$

The left integral is: $$\int (v-1{)}^{-2}dv=-\frac{1}{v-1}$$

The right integral is: $$\frac{1}{2}\int \frac{dx}{x}=\frac{1}{2}\ln |x|+C$$

Combine the results: $$-\frac{1}{v-1}=\frac{1}{2}\ln |x|+C$$

Step 7: Solve for $v$ and Substitute Back

Multiply both sides by $-1$: $$\frac{1}{v-1}=-\frac{1}{2}\ln |x|-C$$

Let $C^{\prime }=-C$: $$\frac{1}{v-1}=-\frac{1}{2}\ln |x|+C^{\prime }$$

Recall that $v=\frac{y}{x}$: $$\frac{1}{\frac{y}{x}-1}=-\frac{1}{2}\ln |x|+C^{\prime }$$

Simplify the left-hand side: $$\frac{x}{y-x}=-\frac{1}{2}\ln |x|+C^{\prime }$$

Let $C^{\prime \prime }=C^{\prime }$ (absorbing constants): $$\frac{x}{y-x}=-\frac{1}{2}\ln |x|+C^{\prime \prime }$$

Step 8: Final Solution

The general solution to the differential equation is: $$\frac{x}{y-x}=-\frac{1}{2}\ln |x|+C$$

Where $C$ is an arbitrary constant.

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b) Solution of the Differential Equation: $(x^2+1)\frac{dy}{dx}+y^2=0$

Step 1: Rewrite the Equation

The given differential equation is: $$(x^2+1)\frac{dy}{dx}+y^2=0$$

Rewrite it to express $\frac{dy}{dx}$: $$(x^2+1)\frac{dy}{dx}=-y^2$$ $$\frac{dy}{dx}=-\frac{y^2}{x^2+1}$$

Step 2: Recognize the Type of Differential Equation

This is a separable equation because it can be written in the form: $$\frac{dy}{y^2}=-\frac{dx}{x^2+1}$$

Step 3: Separate Variables and Integrate

Separate the variables $y$ and $x$: $$\frac{dy}{y^2}=-\frac{dx}{x^2+1}$$

Integrate both sides: $$\int y^{-2}dy=-\int \frac{dx}{x^2+1}$$

The left integral is: $$\int y^{-2}dy=-y^{-1}+C_1$$

The right integral is: $$-\int \frac{dx}{x^2+1}=-{\tan }^{-1}(x)+C_2$$

Combine the results: $$-\frac{1}{y}=-{\tan }^{-1}(x)+C$$

Where $C=C_2-C_1$ is the combined constant of integration.

Step 4: Solve for $y$

Multiply both sides by $-1$: $$\frac{1}{y}={\tan }^{-1}(x)-C$$

Let $C^{\prime }=-C$: $$\frac{1}{y}={\tan }^{-1}(x)+C^{\prime }$$

Solve for $y$: $$y=\frac{1}{{\tan }^{-1}(x)+C^{\prime }}$$

Step 5: Final Solution

The general solution to the differential equation is: $$y=\frac{1}{{\tan }^{-1}(x)+C}$$

Where $C$ is an arbitrary constant.

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Summary of Solutions

For the given options, the solutions are:

1. For $2x^{2}dy=(x^2+y^2)dx$: $$\frac{x}{y-x}=-\frac{1}{2}\ln |x|+C$$

2. For $(x^2+1)\frac{dy}{dx}+y^2=0$: $$y=\frac{1}{{\tan }^{-1}(x)+C}$$