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  • Explain solution for RD Sharma Class 12 Chapter 21 Differential Equation Exercise 21.10 Question 23 maths textbook solution.

Explain solution for RD Sharma Class 12 Chapter 21 Differential Equation Exercise 21.10 Question 23 maths textbook solution.

Answers (1)

Answer : y=\left(\frac{y+1}{y}\right)+C e^{\frac{1}{y}}

Hint : To solve this equation we use e^{\int P d y} formula.

Give : y^{2} \frac{d x}{d y}+x-\frac{1}{y}=0

Solution : \frac{1}{y^{2}}\left[y^{2}\left(\frac{d x}{d y}\right)+x-\frac{1}{y}\right]=0

                \begin{aligned} &=\frac{d y}{d x}+\frac{x^{2}}{y^{2}}-\frac{1}{y^{3}}=0 \\ \end{aligned}

               \begin{aligned} &=\frac{d y}{d x}+\frac{x^{2}}{y^{2}}=\frac{1}{y^{3}} \\ \end{aligned}

               \begin{aligned} &=\frac{d x}{d y}+P(x)=Q \\ \end{aligned}

                \begin{aligned} &P=\frac{1}{y^{\prime}} Q=\frac{1}{y^{3}} \\ \end{aligned}

                \begin{aligned} &I f=e^{\int P d y} \end{aligned}

                \begin{aligned} &=x I f=\int I f Q d y \\ \end{aligned}

                \begin{aligned} &\frac{d x}{d y}+\frac{x}{y^{2}}=\frac{1}{y^{3}} \\ \end{aligned}

                \begin{aligned} &=I f=e^{\int \frac{1}{y^{2}} d y} \\ \end{aligned}

                \begin{aligned} &=e^{-\frac{1}{y}} \\ \end{aligned}

               \begin{aligned} &=x e^{-\frac{1}{y}}=e^{-\frac{1}{y}} \frac{1}{y^{3}} d y \end{aligned}

               \begin{aligned} &=\int \frac{1}{y} e^{-\frac{1}{y}} \frac{1}{y^{2}} d y \\ \end{aligned}

               \begin{aligned} &=-\frac{1}{y}=t \\ \end{aligned}

               \begin{aligned} &=\frac{1}{y^{2}} d y=d t \\ \end{aligned}

               \begin{aligned} &=d y=y^{2} d t \\ \end{aligned}

               \begin{aligned} &=\int t e^{t} d t \\ \end{aligned}

                \begin{aligned} &=x e^{-\frac{1}{y}}=\left[t e^{t}-e^{t}\right]+C \\ \end{aligned}

                \begin{aligned} &=x e^{-\frac{1}{y}}=-t e^{t}+e^{t}+C \\ \end{aligned}

                \begin{aligned} &=x e^{-\frac{1}{y}}=\frac{1}{y} e^{-\frac{1}{y}}+e^{-\frac{1}{y}}+C \\ \end{aligned}

                \begin{aligned} &=x e^{-\frac{1}{y}}=e^{-\frac{1}{y}}\left(\frac{1}{y}+1+\frac{C}{e^{-\frac{1}{y}}}\right) \\ &=x=\frac{1+y}{y}+C e^{\frac{1}{5}} \end{aligned}

             

              

 

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