demo23

Unit 1 ପ୍ରଥମ ଏକକ (ଗଦ୍ୟ)
Chapter 1 ଇତିହାସ	PDF Solution
Chapter 2 ପୁଷ୍ପପୁରରେ ବର୍ଷାବରଣ	PDF Solution
Chapter 3 ତିନି ତୁଣ୍ଡରେ	PDF Solution
Chapter 4 ସ୍ଵାଧୀନ ଦେଶରେ ଶିକ୍ଷାଚିନ୍ତା	PDF Solution

Unit 2 ଦ୍ବିତୀୟ ଏକକ (ପଦ୍ୟ)
Chapter 5 ବଡ଼ପଣ	PDF Solution
Chapter 6 ତପସ୍ବିନୀର ପତ୍ର	PDF Solution
Chapter 7 ବନ୍ଦୀର ବିରହ ବ୍ୟଥା	PDF Solution
Chapter 8 ପିଙ୍ଗଳାର ଅଭିସାର	PDF Solution
Chapter 9 ବାର୍ତା	PDF Solution
Unit 3 ତୃତୀୟ ଏକକ (ଗଳ୍ପ)
Chapter 10 ସଭ୍ୟ ଜମିଦାର	PDF Solution
Chapter 11 ପତାକା ଉତ୍ତୋଳନ	PDF Solution
Chapter 12 ରୂପନାରାୟଣ ସାହା	PDF Solution
Chapter 13 ଆକାଶ କଇଁଛ	PDF Solution

 

Given differential equation is

$(x+y+1)\frac{dy}{dx}=1$

.So,

$(x+y+1)=\frac{dx}{dy}$ $\frac{dx}{dy}–x=y+1$

Which is a linear differential equation in

$x$

.On Comparing it with

$\frac{dx}{dy}+Px=Q$

, We get

$P=–1,Q=y+1$ $\mathrm{IF}=e^{\int Pdy}=e^{\int –dy}$ $=e^{–y}$

So, the integrating factor,

$\mathrm{IF}=e^{–y}$

The complete solution is

$x·\mathrm{IF}=\int Q·\mathrm{IF}dy$ $\Rightarrow x·e^{–y}=\int \left(y+1\right)·e^{–y}dy$ $\Rightarrow x{e}^{–y}=–\left(y+1\right)e^{–y}+\int e^{–y}dy$

[Integration by parts]

$\Rightarrow x{e}^{–y}=–\left(y+1\right)e^{–y}–e^{–y}+c$ $\Rightarrow x{e}^{–y}=–e^{–y}\left(y+2\right)+c$

Which is the required solution of given differential equation.