Solve the following differential equations:

(i) $\dfrac{dy}{dx}=\sqrt{\dfrac{1-y^2}{1-x^2}}$

(ii) $y\,dx+\left(1+x^2 ight) an^{-1}x\,dy=0$

(iii) $\sin\dfrac{dy}{dx}=a,\ \ y(0)=1$

(iv) $\dfrac{dy}{dx}=e^{x+y}+x^3e^y$

(v) $\left(e^y+1 ight)\cos x\,dx+e^y\sin x\,dy=0$

(vi)

Question

Solve the following differential equations:

(i) $\dfrac{dy}{dx}=\sqrt{\dfrac{1-y^2}{1-x^2}}$

(ii) $y\,dx+\left(1+x^2ight)	an^{-1}x\,dy=0$

(iii) $\sin\dfrac{dy}{dx}=a,\ \ y(0)=1$

(iv) $\dfrac{dy}{dx}=e^{x+y}+x^3e^y$

(v) $\left(e^y+1ight)\cos x\,dx+e^y\sin x\,dy=0$

(vi) $\left(y\,dx-x\,dyight)\cot\!\left(\dfrac{x}{y}ight)=ny^2\,dx$

(vii) $\dfrac{dy}{dx}-x\sqrt{25-x^2}=0$

(viii) $x\cos y\,dy=e^x\left(x\log x+1ight)dx$

(ix) $	an y\dfrac{dy}{dx}=\cos(x+y)+\cos(x-y)$

(x) $\dfrac{dy}{dx}=	an^2(x+y)$

Accepted Answer

> [!TLDR]
>
> Each part separates directly (once rearranged), using a standard integral, a $u$-substitution, or a trig identity to combine terms before integrating.

**Step 1. (i) $\dfrac{dy}{dx}=\sqrt{\dfrac{1-y^2}{1-x^2}}$.** Separate: $\dfrac{dy}{\sqrt{1-y^2}}=\dfrac{dx}{\sqrt{1-x^2}}$. Integrate: $\sin^{-1}y=\sin^{-1}x+C$.

**Step 2. (ii) $y\,dx+\left(1+x^2ight)	an^{-1}x\,dy=0$.** Separate: $\dfrac{dy}{y}=-\dfrac{dx}{\left(1+x^2ight)	an^{-1}x}$. Integrate (right side is $-d[\ln(	an^{-1}x)]$ since $\frac{d}{dx}	an^{-1}x=\frac1{1+x^2}$): $\ln|y|=-\ln|	an^{-1}x|+C_1\ \Longrightarrow\ y	an^{-1}x=C$.

**Step 3. (iii) $\sin\dfrac{dy}{dx}=a,\ y(0)=1$.** This means $\dfrac{dy}{dx}=\sin^{-1}a$, a constant. Integrate: $y=x\sin^{-1}a+C$. Apply $y(0)=1$: $C=1$. So $y=x\sin^{-1}a+1$.

**Step 4. (iv) $\dfrac{dy}{dx}=e^{x+y}+x^3e^y=e^y\left(e^x+x^3ight)$.** Separate: $e^{-y}\,dy=\left(e^x+x^3ight)dx$. Integrate: $-e^{-y}=e^x+\dfrac{x^4}{4}+C$.

**Step 5. (v) $\left(e^y+1ight)\cos x\,dx+e^y\sin x\,dy=0$.** Separate: $\dfrac{\cos x}{\sin x}dx=-\dfrac{e^y}{e^y+1}dy\ \Longrightarrow\ \cot x\,dx=-\dfrac{e^y}{e^y+1}dy$. Integrate: $\ln|\sin x|=-\ln\left(e^y+1ight)+C_1\ \Longrightarrow\ \sin x\left(e^y+1ight)=C$.

**Step 6. (vi) $\left(y\,dx-x\,dyight)\cot\!\left(\dfrac{x}{y}ight)=ny^2\,dx$.** Divide both sides by $y^2$ and recognise $d\!\left(\dfrac{x}{y}ight)=\dfrac{y\,dx-x\,dy}{y^2}$: $\cot\!\left(\dfrac{x}{y}ight)d\!\left(\dfrac{x}{y}ight)=n\,dx$. Let $w=\dfrac{x}{y}$: $\cot w\,dw=n\,dx$. Integrate: $\ln|\sin w|=nx+C_1\ \Longrightarrow\ \sin\!\left(\dfrac{x}{y}ight)=Ce^{nx}$.

**Step 7. (vii) $\dfrac{dy}{dx}=x\sqrt{25-x^2}$.** Already separated: $dy=x\sqrt{25-x^2}\,dx$. Let $u=25-x^2,\ du=-2x\,dx$: $\displaystyle\int x\sqrt{25-x^2}\,dx=-\dfrac13(25-x^2)^{3/2}+C$. So $y=-\dfrac13(25-x^2)^{3/2}+C$.

**Step 8. (viii) $x\cos y\,dy=e^x\left(x\ln x+1ight)dx$.** Separate: $\cos y\,dy=e^x\left(\ln x+\dfrac1xight)dx$. Recognise the right side as $d\!\le

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Concept understanding — Variables Separable

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