Question:
The general solution of $\frac{dy}{dx}=\frac{2x-y}{x+2y}$ is given by
The general solution of $\frac{dy}{dx}=\frac{2x-y}{x+2y}$ is given by
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Differential Equations Tip: While this is also a homogeneous differential equation (solvable by substituting $y = vx$), looking for exact differentials like $d(xy) = x\,dy + y\,dx$ is significantly faster!
Updated On: Apr 30, 2026
- $x^{2}-y^{2}-xy=C$
- $x^{2}+y^{2}+xy=C$
- $x^{2}+2y^{2}+y+x=C$
- $2x^{2}+y^{2}+xy+y=C$
- $x^{2}-y^{2}-xy+x=C$
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The Correct Option is A
Solution and Explanation
Concept:
This is a first-order differential equation. It can be solved by rewriting it in differential form ($Mdx + Ndy = 0$) and identifying exact differentials, such as $d(xy) = xdy + ydx$, which allows for direct integration of all terms.
Step 1: Rewrite in differential form.
Cross-multiply to separate the differentials $dx$ and $dy$: $$(x + 2y)dy = (2x - y)dx$$
Step 2: Move all terms to one side of the equation.
Rearrange to group all terms together, setting the equation to zero: $$(2x - y)dx - (x + 2y)dy = 0$$
Step 3: Expand the terms.
Distribute $dx$ and $dy$ to their respective terms: $$2x\,dx - y\,dx - x\,dy - 2y\,dy = 0$$
Step 4: Group to form exact differentials.
Notice the middle terms $-y\,dx - x\,dy$. Factor out the negative sign: $$2x\,dx - (y\,dx + x\,dy) - 2y\,dy = 0$$ Recognize the Product Rule in reverse: $(y\,dx + x\,dy) = d(xy)$. $$2x\,dx - d(xy) - 2y\,dy = 0$$
Step 5: Integrate the entire equation.
Apply the integral operator to each term: $$\int 2x\,dx - \int d(xy) - \int 2y\,dy = \int 0$$ $$x^2 - xy - y^2 = C$$ Hence the correct answer is (A) $x^{2-y^{2}-xy=C$}.
This is a first-order differential equation. It can be solved by rewriting it in differential form ($Mdx + Ndy = 0$) and identifying exact differentials, such as $d(xy) = xdy + ydx$, which allows for direct integration of all terms.
Step 1: Rewrite in differential form.
Cross-multiply to separate the differentials $dx$ and $dy$: $$(x + 2y)dy = (2x - y)dx$$
Step 2: Move all terms to one side of the equation.
Rearrange to group all terms together, setting the equation to zero: $$(2x - y)dx - (x + 2y)dy = 0$$
Step 3: Expand the terms.
Distribute $dx$ and $dy$ to their respective terms: $$2x\,dx - y\,dx - x\,dy - 2y\,dy = 0$$
Step 4: Group to form exact differentials.
Notice the middle terms $-y\,dx - x\,dy$. Factor out the negative sign: $$2x\,dx - (y\,dx + x\,dy) - 2y\,dy = 0$$ Recognize the Product Rule in reverse: $(y\,dx + x\,dy) = d(xy)$. $$2x\,dx - d(xy) - 2y\,dy = 0$$
Step 5: Integrate the entire equation.
Apply the integral operator to each term: $$\int 2x\,dx - \int d(xy) - \int 2y\,dy = \int 0$$ $$x^2 - xy - y^2 = C$$ Hence the correct answer is (A) $x^{2-y^{2}-xy=C$}.
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