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  • Provide solution for RD Sharma maths Class 12 Chapter 21 Differential Equation Exercise Revision Exercise (RE) Question 58 textbook solution.

Provide solution for  RD Sharma maths Class 12 Chapter 21 Differential Equation Exercise Revision Exercise (RE) Question 58 textbook solution.

Answers (1)

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

Hint                : you must know the rules of solving differential equation and integration

Given             :  y^{2}+\left(x+\frac{1}{y}\right) \frac{d y}{d x}=0

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

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

Comparing with, \frac{d x}{d y}+P x=Q

Where  , P=\frac{1}{y^{2}}, Q=\frac{-1}{y^{3}}

Now,

\text { I. } F=e^{\int \frac{1}{y^{2}} d y}=e^{\frac{-1}{y}}

therefore the solution is

\begin{aligned} x \times I . F &=\int \text { I.F } \times Q d y+c \\ x e^{\frac{-1}{y}} &=\int-e^{\frac{-1}{y}} \frac{1}{y^{3}} d y+c \\ x e^{\frac{-1}{y}} &=I+c \end{aligned}

Putting t=\frac{1}{y}, and differentiating both sides

d t=\frac{-1}{y^{2}} d y

Applying integration both side,

\begin{aligned} I &=\int\left(t \times e^{-t}\right) d t \\ I &=t \times \int e^{-t} d t-\int\left(\frac{d t}{d t} \times \int e^{-t} d t\right) d t \\ &=-t e^{-t}+\int e^{-t} d t \\ &=-t e^{-t}-e^{-t} \end{aligned}

Therefore,

\begin{aligned} &I=\frac{-1}{y} e^{\frac{-1}{y}}-e^{\frac{-1}{y}} \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad\left[t=\frac{1}{y}\right] \\ &I=-e^{\frac{-1}{y}}\left(1+\frac{1}{y}\right) \end{aligned}

Hence, required solution is ,

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

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