Question Statement

Solve the differential equation:

$\frac{dy}{dx}+y^2\sin x=0$

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Background and Explanation

This is a first-order ordinary differential equation that can be solved using separation of variables. The equation is separable because we can rearrange terms to isolate all $y$ expressions on one side with $dy$ and all $x$ expressions on the other side with $dx$.

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Solution

Starting with the given equation:

$\frac{dy}{dx}+y^2\sin x=0$

Rearrange to isolate the derivative:

$\frac{dy}{dx}=-y^2\sin x$

Separate the variables by dividing both sides by $y^2$ (where $y\ne 0$) and multiplying by $dx$:

$\frac{dy}{y^2}=-\sin xdx$

Rewrite $\frac{1}{y^2}$ as $y^{-2}$ to prepare for integration:

$y^{-2}dy=-\sin xdx$

Integrate both sides:

$\int y^{-2}dy=\int -\sin xdx$

Evaluate the left side using the power rule:

$\frac{y^{-2+1}}{-2+1}=\frac{y^{-1}}{-1}=-\frac{1}{y}$

Evaluate the right side:

$\int -\sin xdx=\cos x+C$

Equating both results:

$-\frac{1}{y}=\cos x+C$

Multiply both sides by $-1$:

$\frac{1}{y}=-\cos x-C$

Absorb the negative sign into the constant (letting $c=-C$):

$\frac{1}{y}=-\cos x+c$

Alternative approach (as shown in the original derivation):
Starting from $\frac{dy}{y^2}=-\sin xdx$, multiply both sides by $-1$ to obtain $-\frac{dy}{y^2}=\sin xdx$, then integrate:

$-\int y^{-2}dy=\int \sin xdx$

$-\left(\frac{y^{-1}}{-1}\right)=-\cos x+c$

$\frac{1}{y}=-\cos x+c$

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Key Formulas or Methods Used

• Separation of variables: Rearranging $\frac{dy}{dx}=f(x)g(y)$ into $\frac{dy}{g(y)}=f(x)dx$
• Power rule for integration: $\int y^{n}dy=\frac{y^{n+1}}{n+1}$ for $n\ne -1$
• Trigonometric integration: $\int \sin xdx=-\cos x+C$

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Summary of Steps

1. Rearrange the equation to isolate $\frac{dy}{dx}$ on one side
2. Separate variables: divide by $y^2$ and multiply by $dx$ to get $\frac{dy}{y^2}=-\sin xdx$
3. Integrate both sides: $\int y^{-2}dy=\int -\sin xdx$
4. Evaluate to obtain $-\frac{1}{y}=\cos x+C$
5. Solve for the dependent variable: $\frac{1}{y}=-\cos x+c$ (general solution)