Question

cosy dy+ (siny -(x+1)^2)dx=0

Answer 1

Given to solve the differential equation,

$\longrightarrow\cos y\,dy+\left(\sin y-(x+1)^2\right)\,dx=0$

Dividing by $dx,$

$\longrightarrow \cos y\,\dfrac{dy}{dx}+\sin y-(x+1)^2=0$

$\longrightarrow \cos y\,\dfrac{dy}{dx}+\sin y=(x+1)^2$

Put,

$\longrightarrow u=\sin y$

$\longrightarrow \dfrac{du}{dx}=\cos y\,\dfrac{dy}{dx}$

Then,

$\longrightarrow\dfrac{du}{dx}+u=(x+1)^2$

Multiply each term by a function in $x$ only, i.e., say $f(x).$

$\longrightarrow\dfrac{du}{dx}\cdot f(x)+u\cdot f(x)=(x+1)^2\cdot f(x)$

Assume $\dfrac{d}{dx}\left[f(x)\right]=f(x)$ such that,

$\longrightarrow\dfrac{d}{dx}\left[u\cdot f(x)\right]=(x+1)^2\cdot f(x)\quad\quad\dots(1)$

So,

$\longrightarrow\dfrac{d}{dx}\left[f(x)\right]=f(x)$

$\longrightarrow f'(x)=f(x)$

$\longrightarrow\dfrac{f'(x)}{f(x)}=1$

Multiplying by $dx,$

$\longrightarrow\dfrac{f'(x)}{f(x)}\ dx=\ dx$

Integrating,

$\displaystyle\longrightarrow\int\dfrac{f'(x)}{f(x)}\ dx=\int dx$

$\displaystyle\longrightarrow\log|f(x)|=x+k_1$

$\displaystyle\longrightarrow f(x)=e^{x+k_1}$

Taking $e^{k_1}=C_1,$

$\displaystyle\longrightarrow f(x)=C_1\,e^x$

Then (1) becomes,

$\longrightarrow\dfrac{d}{dx}\left[C_1ue^x\right]=C_1(x+1)^2e^x$

$\longrightarrow\dfrac{d}{dx}\left[ue^x\right]=(x+1)^2e^x$

$\longrightarrow d\left[ue^x\right]=(x+1)^2e^x\,dx$

Integrating,

$\displaystyle\longrightarrow ue^x=\int(x+1)^2e^x\,dx$

$\displaystyle\longrightarrow ue^x=e^x\left[(x+1)^2-2(x+1)+2\right]+C_2$

[Note that $d(x+1)=dx.$]

$\displaystyle\longrightarrow ue^x=e^x\left(x^2+1\right)+C_2$

Undoing substitution $u=\sin y$ and taking $C_2=C,$

$\displaystyle\longrightarrow e^x\sin y=e^x\left(x^2+1\right)+C$

$\displaystyle\longrightarrow\sin y=x^2+Ce^{-x}+1$

$\displaystyle\longrightarrow\underline{\underline{y=\sin^{-1}\left(x^2+Ce^{-x}+1\right)}}$

Answer 2

$\huge{\underline{\underline{\mathrm{\red{AnswEr}}}}}$

Given to solve the differential equation,

$\longrightarrow\cos y\,dy+\left(\sin y-(x+1)^2\right)\,dx=0$

Dividing by $dx,$

$\longrightarrow \cos y\,\dfrac{dy}{dx}+\sin y-(x+1)^2=0$

$\longrightarrow \cos y\,\dfrac{dy}{dx}+\sin y=(x+1)^2$

Put,

$\longrightarrow u=\sin y$

$\longrightarrow \dfrac{du}{dx}=\cos y\,\dfrac{dy}{dx}$

Then,

$\longrightarrow\dfrac{du}{dx}+u=(x+1)^2$

Multiply each term by a function in $x$ only, i.e., say $f(x).$

$\longrightarrow\dfrac{du}{dx}\cdot f(x)+u\cdot f(x)=(x+1)^2\cdot f(x)$

Assume $\dfrac{d}{dx}\left[f(x)\right]=f(x)$ such that,

$\longrightarrow\dfrac{d}{dx}\left[u\cdot f(x)\right]=(x+1)^2\cdot f(x)\quad\quad\dots(1)$

So,

$\longrightarrow\dfrac{d}{dx}\left[f(x)\right]=f(x)$

$\longrightarrow f'(x)=f(x)$

$\longrightarrow\dfrac{f'(x)}{f(x)}=1$

Multiplying by $dx,$

$\longrightarrow\dfrac{f'(x)}{f(x)}\ dx=\ dx$

Integrating,

$\displaystyle\longrightarrow\int\dfrac{f'(x)}{f(x)}\ dx=\int dx$

$\displaystyle\longrightarrow\log|f(x)|=x+k_1$

$\displaystyle\longrightarrow f(x)=e^{x+k_1}$

Taking $e^{k_1}=C_1,$

$\displaystyle\longrightarrow f(x)=C_1\,e^x$

Then (1) becomes,

$\longrightarrow\dfrac{d}{dx}\left[C_1ue^x\right]=C_1(x+1)^2e^x$

$\longrightarrow\dfrac{d}{dx}\left[ue^x\right]=(x+1)^2e^x$

$\longrightarrow d\left[ue^x\right]=(x+1)^2e^x\,dx$

Integrating,

$\displaystyle\longrightarrow ue^x=\int(x+1)^2e^x\,dx$

$\displaystyle\longrightarrow ue^x=e^x\left[(x+1)^2-2(x+1)+2\right]+C_2$

[Note that $d(x+1)=dx.$]

$\displaystyle\longrightarrow ue^x=e^x\left(x^2+1\right)+C_2$

Undoing substitution $u=\sin y$ and taking $C_2=C,$

$\displaystyle\longrightarrow e^x\sin y=e^x\left(x^2+1\right)+C$

$\displaystyle\longrightarrow\sin y=x^2+Ce^{-x}+1$

$\displaystyle\longrightarrow\underline{\underline{y=\sin^{-1}\left(x^2+Ce^{-x}+1\right)}}$