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Question: Solve the following expression: \(\left( {{x}^{2}}-y{{x}^{2}} \right)dy+\left( {{y}^{2}}+x{{y}^{2}} ...

Solve the following expression: (x2−yx2)dy+(y2+xy2)dx=0\left( {{x}^{2}}-y{{x}^{2}} \right)dy+\left( {{y}^{2}}+x{{y}^{2}} \right)dx=0

Explanation

Solution

Hint: First we will try to separate the variable of x and y. And then by taking them to opposite sides we will integrate both the sides using the formula for integration ∫dxx=ln⁡x\int{\dfrac{dx}{x}=\ln x}and ∫dxx2=−1x\int{\dfrac{dx}{{{x}^{2}}}=\dfrac{-1}{x}} , then the result that we get will be the final answer.

Complete step-by-step answer:
Let’s start solving the question.
We have been given (x2−yx2)dy+(y2+xy2)dx=0\left( {{x}^{2}}-y{{x}^{2}} \right)dy+\left( {{y}^{2}}+x{{y}^{2}} \right)dx=0
Now after solving it we get,
(x2−yx2)dy=−(y2+xy2)dx x2(1−y)dy=−y2(1+x)dx \begin{aligned} & \left( {{x}^{2}}-y{{x}^{2}} \right)dy=-\left( {{y}^{2}}+x{{y}^{2}} \right)dx \\\ & {{x}^{2}}\left( 1-y \right)dy=-{{y}^{2}}\left( 1+x \right)dx \\\ \end{aligned}
Now we can see that the variables can be separated easily.
After separating the variable x and y we get,
(1−y)dyy2=−(1+x)dxx2\dfrac{\left( 1-y \right)dy}{{{y}^{2}}}=\dfrac{-\left( 1+x \right)dx}{{{x}^{2}}}
Now integrating both the sides we get,
∫(1−y)dyy2=∫−(1+x)dxx2 ∫dyy2−∫dyy=∫−dxx2−∫dxx \begin{aligned} & \int{\dfrac{\left( 1-y \right)dy}{{{y}^{2}}}}=\int{\dfrac{-\left( 1+x \right)dx}{{{x}^{2}}}} \\\ & \int{\dfrac{dy}{{{y}^{2}}}-\int{\dfrac{dy}{y}}=}\int{\dfrac{-dx}{{{x}^{2}}}-\int{\dfrac{dx}{x}}} \\\ \end{aligned}
Now we will use the formula of integration,
∫dxx=ln⁡x\int{\dfrac{dx}{x}=\ln x} and ∫dxx2=−1x\int{\dfrac{dx}{{{x}^{2}}}=\dfrac{-1}{x}}
Using the above two formulas we get,
−1y−ln⁡y=−(−1x)−ln⁡x+C 1y+ln⁡y=−1x+ln⁡x+C \begin{aligned} & \dfrac{-1}{y}-\ln y=-\left( \dfrac{-1}{x} \right)-\ln x+C \\\ & \dfrac{1}{y}+\ln y=\dfrac{-1}{x}+\ln x+C \\\ \end{aligned}
Hence, this is the equation that we get after solving the differential equation.

Note: Here we can see that we have got a constant C in our equation, students might be confused about what it is. C is called the integration constant and to find it’s value we must know at least one point on the equation. As in the question no such information is given so, we can’t determine the value of C. And the two formula ∫dxx=ln⁡x\int{\dfrac{dx}{x}=\ln x} and ∫dxx2=−1x\int{\dfrac{dx}{{{x}^{2}}}=\dfrac{-1}{x}} must be kept in mind while solving this question.