Q13 E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

In Problems 9-13, determine whether the given relation is an implicit solution to the given differential equation. Assume that the relationship implicitly defines y as a function of x and use implicit differentiation.

$\mathbf{sin} \mathbf{y}\mathbf{+}\mathbf{xy}\mathbf{-}{\mathbf{x}}^{\mathbf{3}}\mathbf{=}\mathbf{2}$, $\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{=}\frac{\mathbf{6}\mathbf{xy}\mathbf{\text{'}}\mathbf{+}{\left(\mathbf{y}\mathbf{\text{'}}\right)}^{\mathbf{3}}\mathbf{sin} \mathbf{y}\mathbf{-}\mathbf{2}{\left(\mathbf{y}\mathbf{\text{'}}\right)}^{\mathbf{2}}}{\mathbf{3}{\mathbf{x}}^{\mathbf{2}}\mathbf{-}\mathbf{y}}$

Step-by-Step Solution

Answer

The given relation is an implicit solution to the given differential equation.

Step1: Differentiating the given relation

As, in the given relation $\sin \mathrm{y}+\mathrm{xy}-{\mathrm{x}}^3=2$, y is defined implicitly as the function of x, so by using implicit differentiation, we will differentiate the given relation concerning x,

 $$\begin{gathered} \frac{\mathrm{d}}{\mathrm{dx}}\left(\sin \mathrm{y}+\mathrm{xy}-{\mathrm{x}}^3\right)=\frac{\mathrm{d}}{\mathrm{dx}}\left(2\right) \\ \cos \mathrm{y}\frac{\mathrm{dy}}{\mathrm{dx}}+\left(\mathrm{x}\frac{\mathrm{dy}}{\mathrm{dx}}+\mathrm{y}\right)-3{\mathrm{x}}^2=0 \\ \mathrm{y}\text{'}\times \mathrm{cosy}+\mathrm{xy}\text{'}+\mathrm{y}-3{\mathrm{x}}^2=0······\left(1\right) \\ \left(\cos \mathrm{y}+\mathrm{x}\right)\mathrm{y}\text{'}+\mathrm{y}-3{\mathrm{x}}^2=0 \\ \mathrm{y}\text{'}=\frac{3{\mathrm{x}}^2-\mathrm{y}}{\cos \mathrm{y}+\mathrm{x}} \end{gathered}$$

  

Now, differentiating (1) concerning x,

$$\begin{gathered} \left(-\sin \mathrm{y}\right)\times {\left(\mathrm{y}\text{'}\right)}^2+\mathrm{y}\text{'}\text{'}\times \left(\cos \mathrm{y}\right)+\left(\mathrm{xy}\text{'}\text{'}+\mathrm{y}\text{'}\right)+\mathrm{y}\text{'}-6\mathrm{x}=0 \\ \left(-\sin \mathrm{y}\right)\times {\left(\mathrm{y}\text{'}\right)}^2+\mathrm{y}\text{'}\text{'}\times \left(\cos \mathrm{y}\right)+\mathrm{xy}\text{'}\text{'}+2\mathrm{y}\text{'}-6\mathrm{x}=0 \\ \left(\cos \mathrm{y}+\mathrm{x}\right)\mathrm{y}\text{'}\text{'}-{\left(\mathrm{y}\text{'}\right)}^2\sin \mathrm{y}+2\mathrm{y}\text{'}-6\mathrm{x}=0 \\ \left(\cos \mathrm{y}+\mathrm{x}\right)\mathrm{y}\text{'}\text{'}={\left(\mathrm{y}\text{'}\right)}^2\sin \mathrm{y}-2\mathrm{y}\text{'}+6\mathrm{x} \\ \mathrm{y}\text{'}\text{'}=\frac{{\left(\mathrm{y}\text{'}\right)}^2\sin \mathrm{y}-2\mathrm{y}\text{'}+6\mathrm{x}}{\left(\cos \mathrm{y}+\mathrm{x}\right)} \end{gathered}$$

Step 2: Simplification of the differential equation obtained in step1.

Multiplying and dividing y' in R.H.S. (Right-hand side) of the equation obtained in step 1,

$$\begin{gathered} \mathrm{y}\text{'}\text{'}=\frac{\left({\left(\mathrm{y}\text{'}\right)}^2\sin \mathrm{y}-2\mathrm{y}\text{'}+6\mathrm{x}\right)\times \mathrm{y}\text{'}}{\left(\cos \mathrm{y}+\mathrm{x}\right)\times \mathrm{y}\text{'}} \\ \mathrm{y}\text{'}\text{'}=\frac{{\left(\mathrm{y}\text{'}\right)}^3\sin \mathrm{y}-2{\left(\mathrm{y}\text{'}\right)}^2+6\mathrm{xy}\text{'}}{\left(\cos \mathrm{y}+\mathrm{x}\right)\times \mathrm{y}\text{'}} \end{gathered}$$

 

Putting the value of y' from step 1,

$$\begin{gathered} \mathrm{y}\text{'}\text{'}=\frac{\left({\left(\mathrm{y}\text{'}\right)}^3\sin \mathrm{y}-2{\left(\mathrm{y}\text{'}\right)}^2+6\mathrm{xy}\text{'}\right)\times \left(\cos \mathrm{y}+\mathrm{x}\right)}{\left(\cos \mathrm{y}+\mathrm{x}\right)\times \left(3{\mathrm{x}}^2-\mathrm{y}\right)} \\ \mathrm{y}\text{'}\text{'}=\frac{6\mathrm{xy}\text{'}+{\left(\mathrm{y}\text{'}\right)}^3\sin \mathrm{y}-2{\left(\mathrm{y}\text{'}\right)}^2}{\left(3{\mathrm{x}}^2-\mathrm{y}\right)} \end{gathered}$$

Which is identical to the given differential equation. 

 

Thus, the relation $\mathbf{sin}\mathbf{ }\mathbf{y}\mathbf{+}\mathbf{xy}\mathbf{-}{\mathbf{x}}^{\mathbf{3}}\mathbf{=}\mathbf{2}\mathbf{,}$ is an implicit solution to the differential equation $\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{=}\frac{\mathbf{6}\mathbf{xy}\mathbf{\text{'}}\mathbf{+}{\left(\mathbf{y}\mathbf{\text{'}}\right)}^{\mathbf{3}}\mathbf{sin} \mathbf{y}\mathbf{-}\mathbf{2}{\left(\mathbf{y}\mathbf{\text{'}}\right)}^{\mathbf{2}}}{\mathbf{3}{\mathbf{x}}^{\mathbf{2}}\mathbf{-}\mathbf{y}}$.