Question Statement

Solve the differential equation:

$(x^2+2y^2)dx=xydy$

subject to the initial condition $y(1)=1$.

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Background and Explanation

This is a homogeneous differential equation, where every term in the numerator and denominator has the same degree (degree 2 in this case). Such equations are solved using the substitution $y=ux$ to transform them into separable differential equations.

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Solution

First, rewrite the differential equation in standard form $\frac{dy}{dx}=f(x,y)$:

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

Identifying the Equation Type

This is a homogeneous differential equation because the function $f(x,y)=\frac{x^2+2y^2}{xy}$ can be expressed as a function of $\frac{y}{x}$ alone. We solve this using the substitution $y=ux$.

Substitution and Differentiation

Let $y=ux$, where $u$ is a function of $x$. Differentiating with respect to $x$ using the product rule:

$\frac{dy}{dx}=x\frac{du}{dx}+u$

Substituting into equation (1):

$x\frac{du}{dx}+u=\frac{x^2+2(ux{)}^2}{x(ux)}$

Simplify the right-hand side by factoring out $x^2$:

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

Separation of Variables

Isolate the derivative term:

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

Separate the variables by moving all $u$ terms to the left and $x$ terms to the right:

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

Integration

Integrate both sides:

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

Notice that the numerator on the left is half the derivative of the denominator. Multiply and divide by 2:

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

This yields:

$\frac{1}{2}\ln (1+u^2)=\ln x+\ln c$

where $\ln c$ is the constant of integration. Using logarithm properties to combine terms:

$\ln (1+u^2{)}^{1/2}=\ln (cx)$

Exponentiating both sides to remove the logarithms:

$(1+u^2{)}^{1/2}=cx$

Back-Substitution

Since $y=ux$, we have $u=\frac{y}{x}$. Substituting back into equation (2):

${\left(1+\frac{y^2}{x^2}\right)}^{1/2}=cx$

Combine the terms inside the parentheses:

${\left(\frac{x^2+y^2}{x^2}\right)}^{1/2}=cx$

Simplify the left side:

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

Multiply both sides by $x$:

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

Applying the Initial Condition

Given $y(1)=1$, substitute $x=1$ and $y=1$ into equation (3):

$\sqrt{1^2+1^2}=c(1{)}^2$

$\sqrt{2}=c$

Final Solution

Substitute $c=\sqrt{2}$ back into equation (3):

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

This can also be written as $x^2+y^2=2x^4$ or $y^2=x^2(2x^2-1)$.

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Key Formulas or Methods Used

• Homogeneous Differential Equation: Identified by $\frac{dy}{dx}=f\left(\frac{y}{x}\right)$, solved using substitution $y=ux$
• Product Rule: $\frac{d}{dx}(ux)=x\frac{du}{dx}+u$
• Separation of Variables: Rearranging to form $\frac{u}{1+u^2}du=\frac{dx}{x}$
• Integration Formula: $\int \frac{f^{\prime }(x)}{f(x)}dx=\ln |f(x)|+C$
• Logarithm Properties: $a\ln b=\ln (b^a)$ and $\ln a+\ln b=\ln (ab)$

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Summary of Steps

1. Rewrite as $\frac{dy}{dx}=\frac{x^2+2y^2}{xy}$ and identify as homogeneous
2. Substitute $y=ux$ and differentiate: $\frac{dy}{dx}=x\frac{du}{dx}+u$
3. Substitute into the equation and simplify to $x\frac{du}{dx}=\frac{1+u^2}{u}$
4. Separate variables: $\frac{u}{1+u^2}du=\frac{dx}{x}$
5. Integrate both sides: $\frac{1}{2}\ln (1+u^2)=\ln x+\ln c$
6. Simplify to $(1+u^2{)}^{1/2}=cx$
7. Substitute back $u=\frac{y}{x}$ to obtain $\sqrt{x^2+y^2}=c{x}^2$
8. Apply $y(1)=1$ to find $c=\sqrt{2}$
9. State final solution: $\sqrt{x^2+y^2}=\sqrt{2}{x}^2$