Question

(i) Solve ${{x}^{2}}\dfrac{dy}{dx}={{x}^{2}}+xy+{{y}^{2}}$
(ii) Solve ${{y}^{2}}dx+\left( xy+{{x}^{2}} \right)dy=0$
(iii) Solve ${{x}^{2}}ydx-\left( {{x}^{3}}+{{y}^{3}} \right)dy=0$
(iv) Solve $\left( {{x}^{2}}-{{y}^{2}} \right)dx+2xydy=0$

Answer 1

Hint: While solving this type of questions we should simplify the given equations from the questions and perform required integrations and differentiations on them using the formulae like $\int{{{x}^{n}}dx}=\dfrac{{{x}^{n+1}}}{n+1}$ when $n\ne -1$ , when $n=-1$ we will apply the formula $\int{{{x}^{-1}}dx=\log x}$ . Take each part and rearrange the terms such that dx is on one side and dy is on one side of the equation. Then integrate both sides by applying the above mentioned formula where required and simplify to get the answer.

Complete answer:
For answering this question we will solve each part step by step.
(i) For the first part of the question let us simplify the given equation
${{x}^{2}}\dfrac{dy}{dx}={{x}^{2}}+xy+{{y}^{2}}$
After rearranging we will get ${{x}^{2}}dy=\left( {{x}^{2}}+xy+{{y}^{2}} \right)dx$
By performing integration on both sides we will get $\int{{{x}^{2}}dy}=\int{\left( {{x}^{2}}+xy+{{y}^{2}} \right)dx}$
By applying the formulae $\int{{{x}^{n}}dx}=\dfrac{{{x}^{n+1}}}{n+1}$ for the above equation we get
${{x}^{2}}y=\dfrac{{{x}^{3}}}{3}+\dfrac{{{x}^{2}}}{2}y+x{{y}^{2}}$
By simplifying this we will get
$\begin{align}
  & 6{{x}^{2}}y=2{{x}^{3}}+3{{x}^{2}}y+6x{{y}^{2}} \\
 & \Rightarrow 2{{x}^{3}}-3{{x}^{2}}y+6x{{y}^{2}}=0 \\
\end{align}$
The result is $2{{x}^{3}}-3{{x}^{2}}y+6x{{y}^{2}}=0$
(ii) For the second part of the question let us simplify the given equation
${{y}^{2}}dx+\left( xy+{{x}^{2}} \right)dy=0$
After rearranging we will get ${{y}^{2}}dx=-\left( xy+{{x}^{2}} \right)dy$
By performing integration on both sides we will get $\int{{{y}^{2}}dx}=\int{-\left( xy+{{x}^{2}} \right)dy}$
By applying the formulae $\int{{{x}^{n}}dx}=\dfrac{{{x}^{n+1}}}{n+1}$ for the above equation we get
${{y}^{2}}x=-\left( x\dfrac{{{y}^{2}}}{2}+{{x}^{2}}y \right)$
By simplifying this we will get
$\begin{align}
  & {{y}^{2}}x=-\left( x\dfrac{{{y}^{2}}}{2}+{{x}^{2}}y \right) \\
 & \Rightarrow 2{{y}^{2}}x+x{{y}^{2}}+2{{x}^{2}}y=0 \\
 & \Rightarrow 3x{{y}^{2}}+2{{x}^{2}}y=0 \\
\end{align}$
The result is $3x{{y}^{2}}+2{{x}^{2}}y=0$
(iii) For the third part of the question let us simplify the given equation
${{x}^{2}}ydx-\left( {{x}^{3}}+{{y}^{3}} \right)dy=0$
After rearranging we will get ${{x}^{2}}ydx=\left( {{x}^{3}}+{{y}^{3}} \right)dy$
By performing integration on both sides we will get $\int{{{x}^{2}}ydx}=\int{\left( {{x}^{3}}+{{y}^{3}} \right)dy}$
By applying the formulae $\int{{{x}^{n}}dx}=\dfrac{{{x}^{n+1}}}{n+1}$ for the above equation we get
$\dfrac{{{x}^{3}}}{3}y={{x}^{3}}y+\dfrac{{{y}^{4}}}{4}$
By simplifying this we will get
$\begin{align}
  & \dfrac{{{x}^{3}}}{3}y={{x}^{3}}y+\dfrac{{{y}^{4}}}{4} \\
 & \Rightarrow 4{{x}^{3}}y=12{{x}^{3}}y+3{{y}^{4}} \\
\end{align}$
The result is $3{{y}^{4}}-8{{x}^{3}}y=0$
(iv) For the fourth part of the question let us simplify the given equation
$\left( {{x}^{2}}-{{y}^{2}} \right)dx+2xydy=0$
After rearranging we will get $\left( {{x}^{2}}-{{y}^{2}} \right)dx=-2xydy$
By performing integration on both sides we will get $\int{\left( {{x}^{2}}-{{y}^{2}} \right)dx}=\int{-2xydy}$
By applying the formulae $\int{{{x}^{n}}dx}=\dfrac{{{x}^{n+1}}}{n+1}$ for the above equation and simplifying we get
$\begin{align}
  & \int{\left( {{x}^{2}}-{{y}^{2}} \right)dx}=\int{-2xydy} \\
 & \dfrac{{{x}^{3}}}{3}-x{{y}^{2}}=-2x\dfrac{{{y}^{2}}}{2} \\
 & \Rightarrow \dfrac{{{x}^{3}}}{3}=-x{{y}^{2}}+x{{y}^{2}} \\
 & \Rightarrow \dfrac{{{x}^{3}}}{3}=0 \\
 & \Rightarrow x=0 \\
\end{align}$
The result is $x=0$
Hence, we conclude that
(i) ${{x}^{2}}\dfrac{dy}{dx}={{x}^{2}}+xy+{{y}^{2}}$
$\Rightarrow 2{{x}^{3}}-3{{x}^{2}}y+6x{{y}^{2}}=0$
(ii) ${{y}^{2}}dx+\left( xy+{{x}^{2}} \right)dy=0$
$\Rightarrow 3x{{y}^{2}}+2{{x}^{2}}y=0$
(iii) ${{x}^{2}}ydx-\left( {{x}^{3}}+{{y}^{3}} \right)dy=0$
$\Rightarrow 3{{y}^{4}}-8{{x}^{3}}y=0$
(iv) $\left( {{x}^{2}}-{{y}^{2}} \right)dx+2xydy=0$
$\Rightarrow x=0$

Note:
While solving the questions of above type we should remember that $\int{{{x}^{n}}dx}=\dfrac{{{x}^{n+1}}}{n+1}$formulae applies only when $n\ne 1$. For $n=-1$ it will be $\int{{{x}^{-1}}dx=\log x}$. When we integrate an expression with respect to $y$ then $x$ will be considered as a constant and we take it out from the integration as it is constant, same can be done in the case of differentiation. The first step of rearranging the terms must be done carefully such that the overall equation doesn’t change, i.e. be careful while opening the brackets and transposing terms. We just have to take dx and dy terms together and not x and y terms together.