Question

Find the solution of $\dfrac{dy}{dx}=1-x\left( y-x \right)-{{x}^{3}}{{\left( y-x \right)}^{3}}$.
A. ${{\left( y-x \right)}^{2}}\left( {{x}^{2}}+1+c{{x}^{2}} \right)=1$
B. ${{\left( y-x \right)}^{2}}\left( {{x}^{2}}+1+c{{e}^{{{x}^{2}}}} \right)=1$
C. ${{\left( y-x \right)}^{2}}\left( {{x}^{2}}-1+c{{x}^{2}} \right)=1$
D. ${{\left( y-x \right)}^{2}}\left( -{{x}^{2}}-1+c{{e}^{{{x}^{2}}}} \right)=1$

Answer 1

We first try to form the differential of ${{e}^{-{{x}^{2}}}}{{\left( y-x \right)}^{-2}}$. The main equation will be divided with ${{\left( y-x \right)}^{3}}$. On the left side we will get the differential form of chain rule of ${{e}^{-{{x}^{2}}}}{{\left( y-x \right)}^{-2}}$. Then we need to find the integral of right-side function of $2{{x}^{3}}{{e}^{-{{x}^{2}}}}dx$. At the end we find the equation similar to the options given.

Complete step-by-step answer:
We have been given the differential equation of $\dfrac{dy}{dx}=1-x\left( y-x \right)-{{x}^{3}}{{\left( y-x \right)}^{3}}$.
We try to form a differential form of $\left( y-x \right)$.
$\begin{aligned} & \dfrac{dy}{dx}=1-x\left( y-x \right)-{{x}^{3}}{{\left( y-x \right)}^{3}} \\\ & \Rightarrow \dfrac{dy}{dx}-1=-x\left( y-x \right)\left[ 1+{{x}^{2}}{{\left( y-x \right)}^{2}} \right] \\\ & \Rightarrow \dfrac{dy-dx}{dx}=-x\left( y-x \right)\left[ 1+{{x}^{2}}{{\left( y-x \right)}^{2}} \right] \\\ & \Rightarrow \dfrac{d\left( y-x \right)}{dx}+x\left( y-x \right)\left[ 1+{{x}^{2}}{{\left( y-x \right)}^{2}} \right]=0 \\\ \end{aligned}$
The differentials and also the equation in the form of $\left( y-x \right)$.
We divide the whole equation with ${{\left( y-x \right)}^{3}}$.

& \dfrac{d\left( y-x \right)}{dx}+x\left( y-x \right)\left[ 1+{{x}^{2}}{{\left( y-x \right)}^{2}} \right]=0 \\\ & \Rightarrow \dfrac{d\left( y-x \right)}{{{\left( y-x \right)}^{3}}dx}+\dfrac{x\left( y-x \right)}{\left( y-x \right)}\left[ \dfrac{1+{{x}^{2}}{{\left( y-x \right)}^{2}}}{{{\left( y-x \right)}^{2}}} \right]=0 \\\ & \Rightarrow \dfrac{d\left( y-x \right)}{{{\left( y-x \right)}^{3}}}+x\left[ {{x}^{2}}+\dfrac{1}{{{\left( y-x \right)}^{2}}} \right]dx=0 \\\ \end{aligned}$$ Now we break the differential equation in the differential form of the multiplication of ${{e}^{-{{x}^{2}}}}{{\left( y-x \right)}^{-2}}$. $$\begin{aligned} & \dfrac{d\left( y-x \right)}{{{\left( y-x \right)}^{3}}}+x\left[ {{x}^{2}}+\dfrac{1}{{{\left( y-x \right)}^{2}}} \right]dx=0 \\\ & \Rightarrow \dfrac{d\left( y-x \right)}{{{\left( y-x \right)}^{3}}}+\dfrac{xdx}{{{\left( y-x \right)}^{2}}}=-{{x}^{3}}dx \\\ \end{aligned}$$ We multiply the term $-2{{e}^{-{{x}^{2}}}}$ both sides and also find the differentiation of ${{e}^{-{{x}^{2}}}}{{\left( y-x \right)}^{-2}}$. $$\begin{aligned} & \dfrac{d\left( y-x \right)}{{{\left( y-x \right)}^{3}}}+\dfrac{xdx}{{{\left( y-x \right)}^{2}}}=-{{x}^{3}}dx \\\ & \Rightarrow {{e}^{-{{x}^{2}}}}\dfrac{-2d\left( y-x \right)}{{{\left( y-x \right)}^{3}}}-\dfrac{2x{{e}^{-{{x}^{2}}}}dx}{{{\left( y-x \right)}^{2}}}=2{{x}^{3}}{{e}^{-{{x}^{2}}}}dx.........(i) \\\ \end{aligned}$$ Now the left-hand side is the differential form of ${{e}^{-{{x}^{2}}}}{{\left( y-x \right)}^{-2}}$. We find the form by going differentiation side. We are finding the value of $d\left[ {{e}^{-{{x}^{2}}}}{{\left( y-x \right)}^{-2}} \right]$. We have chain rule to apply $d\left( xy \right)=ydx+xdy$. $\begin{aligned} & d\left[ {{e}^{-{{x}^{2}}}}{{\left( y-x \right)}^{-2}} \right] \\\ & ={{e}^{-{{x}^{2}}}}\left[ d{{\left( y-x \right)}^{-2}} \right]+{{\left( y-x \right)}^{-2}}d\left( {{e}^{-{{x}^{2}}}} \right) \\\ & ={{e}^{-{{x}^{2}}}}\dfrac{-2d\left( y-x \right)}{{{\left( y-x \right)}^{3}}}-\dfrac{2x{{e}^{-{{x}^{2}}}}dx}{{{\left( y-x \right)}^{2}}} \\\ \end{aligned}$ Now we can change the equation form of (i) in the differential form $$\begin{aligned} & {{e}^{-{{x}^{2}}}}\dfrac{-2d\left( y-x \right)}{{{\left( y-x \right)}^{3}}}-\dfrac{2x{{e}^{-{{x}^{2}}}}dx}{{{\left( y-x \right)}^{2}}}=2{{x}^{3}}{{e}^{-{{x}^{2}}}}dx \\\ & \Rightarrow d\left[ {{e}^{-{{x}^{2}}}}{{\left( y-x \right)}^{-2}} \right]=2{{x}^{3}}{{e}^{-{{x}^{2}}}}dx \\\ \end{aligned}$$ We also have to form the differential of the right-hand side $$2{{x}^{3}}{{e}^{-{{x}^{2}}}}dx$$. $$2{{x}^{3}}{{e}^{-{{x}^{2}}}}dx=\left( -2x \right)\left( -{{x}^{2}} \right){{e}^{-{{x}^{2}}}}dx$$. Let $$-{{x}^{2}}=z\Rightarrow d\left( -{{x}^{2}} \right)=dz$$. Taking differentials, we get $$\begin{aligned} & d\left( -{{x}^{2}} \right)=dz \\\ & \Rightarrow \left( -2x \right)dx=dz \\\ \end{aligned}$$ Replacing the values, we get $$\left( -2x \right)\left( -{{x}^{2}} \right){{e}^{-{{x}^{2}}}}dx=z{{e}^{z}}dz$$. So, the equation becomes $$d\left[ {{e}^{-{{x}^{2}}}}{{\left( y-x \right)}^{-2}} \right]=z{{e}^{z}}dz$$. Taking integration both sides we get $$\begin{aligned} & \int{d\left[ {{e}^{-{{x}^{2}}}}{{\left( y-x \right)}^{-2}} \right]}=\int{z{{e}^{z}}dz} \\\ & \Rightarrow {{e}^{-{{x}^{2}}}}{{\left( y-x \right)}^{-2}}=z{{e}^{z}}-{{e}^{z}}+c \\\ \end{aligned}$$ Here, c is the integral constant. We also need to replace the value of z. $$\begin{aligned} & {{e}^{-{{x}^{2}}}}{{\left( y-x \right)}^{-2}}=z{{e}^{z}}-{{e}^{z}}+c \\\ & \Rightarrow {{e}^{-{{x}^{2}}}}{{\left( y-x \right)}^{-2}}=\left( -{{x}^{2}}-1 \right){{e}^{-{{x}^{2}}}}+c \\\ & \Rightarrow {{\left( y-x \right)}^{-2}}=\left( -{{x}^{2}}-1 \right)+c{{e}^{{{x}^{2}}}} \\\ & \Rightarrow \dfrac{1}{{{\left( y-x \right)}^{2}}}=\left( c{{e}^{{{x}^{2}}}}-{{x}^{2}}-1 \right) \\\ & \Rightarrow {{\left( y-x \right)}^{2}}\left( -{{x}^{2}}-1+c{{e}^{{{x}^{2}}}} \right)=1 \\\ \end{aligned}$$ **So, the correct answer is “Option D”.** **Note:** If we don’t take a differential then we have to take too many variables to compensate for the form and that will become very tough to solve after a time. That’s why the left-hand side has to be solved with the help of differentials.