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Please solve RD Sharma maths Class 12 Chapter 21 Differential Equation Exercise Revision Exercise (RE) Question 66 Subquestion (xiii) textbook solution.

Answers (1)

Answer : x+y+1=ce^{y}

Hint : you integrate by applying integration of  x^{n}

Given : (x+y) \frac{d y}{d x}=1

Solution : Put in the form of \frac{dy}{dx}+Py=Q

(x+y) \frac{d y}{d x}=1

Divide by x+y

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

\frac{d x}{d y}=x+y

\frac{d x}{d y}+(-x)=y                                     .....(i)

Find P_{1} and Q_{1}

Comparing (i)

\begin{aligned} &\frac{d y}{d x}+P_{1}=Q_{1} \\ &P_{1}=-1 \quad, Q_{1}=y \end{aligned}

Find I.F

\begin{aligned} &\text { I.F }=e^{\int P_{1} d y} \\ \end{aligned}

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

x \times I . F=\int Q_{1} \times I . F d y+c

Putting value x e^{-y}=\int y \times e^{-y} d y+c                     ......(ii)

\begin{aligned} &\text { Let } I=\int y e^{-y} d y \\ \end{aligned}

\begin{aligned} &=y \int e^{-y} d y-\int\left(\frac{d}{d y} y \int e^{-y} d y\right) d y \\ \end{aligned}

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

\begin{aligned} &=-y e^{-y}+\int e^{-y} d y \\ \end{aligned}

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

Put value of I in (ii)

\begin{aligned} &x e^{-y}=\int y e^{-y} d y+c \\ &x e^{-y}=-y e^{-y}-e^{-y}+c \end{aligned}

Divide by e^{-y}

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

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