Question

The general solution of the differential equation $\left( {{y}^{2}}+{{x}^{3}} \right)dx-xydy=0\left( x\ne 0 \right)$ is:
(where $c$ is a constant of integration).$$$$
A.{{y}^{2}}+2{{x}^{3}}+c{{x}^{2}}=0$$$$$ B. {{y}^{2}}+2{{x}^{3}}+c{{x}^{3}}=0 C. ${{y}^{2}}-2{{x}^{3}}+c{{x}^{2}}=0
D. ${{y}^{2}}-2{{x}^{3}}+c{{x}^{3}}=0$$$$$

Answer 1

Convert the given differential equation to linear form ${{y}^{'}}+yP\left( x \right)=Q\left( x \right)$ where $P\left( x \right)$ and $Q\left( x \right)$ are real differentiable functions, $C$is a real constant of integration. The solution of the linear differential equation is $y{{e}^{\int{P\left( x \right)}}}=\int{Q\left( x \right)}{{e}^{\int{P\left( x \right)}}}dx+C$ where we use the integrating factor (an expression which multiplied to facilitate integration. ) .

Complete step by step answer:
The order of a differential equation is the highest time derivative in the equation where the degree of a differential equation is the highest power of any variable. If the highest power of a variable and its derivative is 1 then the differential equation of is linear in that variable.$$$$
We know that the solution of first order linear differential equation of the standard form ${{y}^{'}}+yP\left( x \right)=Q\left( x \right)$ is $y{{e}^{\int{P\left( x \right)}}}=\int{Q\left( x \right)}{{e}^{\int{P\left( x \right)}}}dx+C$ where $P\left( x \right)$ and $Q\left( x \right)$ are real differentiable functions, $C$is a real constant of integration and the integrating factor is $\text{IF}={{e}^{\int{P\left( x \right)}dx}}$
The given differential equation is ,
$\left( {{y}^{2}}+{{x}^{3}} \right)dx-xydy=0\left( x\ne 0 \right)..\left( 1 \right)$
We have to convert the above equation to the standard form but we can see that the differential equation is not linear as the variable $y$ is with power 2. So let us substitute ${{y}^{2}}=z$ then by differentiation we get $y\dfrac{dy}{dx}=\dfrac{1}{2}\dfrac{dz}{dx}$.So let us use it to transform the equation to the standard form.

& \left( {{y}^{2}}+{{x}^{3}} \right)dx-xydy=0 \\\ & \Rightarrow {{y}^{2}}+{{x}^{3}}=xy\dfrac{dy}{dx} \\\ & \Rightarrow z+{{x}^{3}}=x\left( \dfrac{1}{2} \right)\left( \dfrac{dz}{dx} \right) \\\ & \Rightarrow \dfrac{dz}{dx}-\dfrac{2}{x}z=2{{x}^{2}} \\\ \end{aligned}$$ Now the above result is a linear differential equation of the standard form where $P\left( x \right)=\dfrac{-2}{x}$ and $Q\left( x \right)=2{{x}^{2}}$ . We find the integrating factor by using integration formula $\int{\dfrac{1}{x}dx=\ln \left| x \right|}$ and the identity ${{e}^{{{\log }_{e}}a}}=a$ . So $\text{IF}={{e}^{\int{\dfrac{\left( -2 \right)}{x}dx}}}={{e}^{\ln \dfrac{1}{{{x}^{2}}}}}=\dfrac{1}{{{x}^{2}}}$ and the solution of the differential equation is given by $$$$ $$\begin{aligned} & z{{e}^{\int{P\left( x \right)}}}=\int{Q\left( x \right)}{{e}^{\int{P\left( x \right)}}}dx+c \\\ & \Rightarrow z\cdot \dfrac{1}{{{x}^{2}}}=\int{2{{x}^{2}}\cdot }\dfrac{1}{{{x}^{2}}}+c \\\ & \Rightarrow z=2{{x}^{3}}+c{{x}^{2}} \\\ & \Rightarrow {{y}^{2}}-2{{x}^{3}}-c{{x}^{2}}=0\left( \because z={{y}^{2}} \right) \\\ \end{aligned}$$ **So, the correct answer is “Option D”.** **Note:** The integrating factor converts an non-exact differential equation of to exact differential equation of the type $Mdx+Ndy=0$ with condition $\dfrac{\partial M}{\partial y}=\dfrac{\partial N}{\partial x}$ where $M,N$ are functions of $x$. The question can also be framed to check whether another point satisfies the solution or not.