Question

Solve the following differential equations: (i) (𝑦 3 + 1) 𝑑𝑦 𝑑𝑥 − 𝑥𝑦 2 = 𝑥, (ii) 𝑥𝑦 2 𝑑𝑦 𝑑𝑥 = 𝑥 3 − 𝑦 3, (iii) (1 + sin 𝑥) 𝑑𝑦 𝑑𝑥 + 𝑦 cos 𝑥 = tan 𝑥, (iv) 𝑥 𝑑𝑦 𝑑𝑥 − 5𝑦 = 𝑥 7.

Answer 1

(i)

$(y^{3}+1)\frac{dy}{dx}=x+xy^{2}$

$(y^{3}+1)\frac{dy}{dx}=x(1+y^{2})$

$\frac{1+y^{3}}{1+y^{2}}dy=xdx$

Integrate on both sides

$\int \frac{1+y^{3}}{1+y^{2}}dy=\int xdx$

Let $1+y^{2}$ =t

2ydy=dt

ydy=$\frac{dt}{2}$

$y^{2}=(t-1)$

$\int \frac{1}{1+y^{2}}dy+\int \frac{y^{3}}{1+y^{2}}=\int xdx$

$tan^{-1}(y)+\frac{1}{2}\int \frac{(t-1)}{t}dt=\frac{x^{2}}{2}+C$

$tan^{-1}(y)+\frac{1}{2}\int (1-\frac{1}{t})dt=\frac{x^{2}}{2}+C$

$tan^{-1}(y)+\frac{1}{2}(t-ln\ t)=\frac{x^{2}}{2}+C$

Putting the values of t answer becomes;

$tan^{-1}(y)+\frac{1}{2}[(1+y^{2})-ln(1+y^{2})]=\frac{x^{2}}{2}+C$

(ii)

$xy^{2}\frac{dy}{dx}=x^{3}-y^{3}$

$\frac{dy}{dx}=\frac{x^{3}-y^{3}}{xy^{2}}$

On dividing the numerator and denominator by x³

$\frac{dy}{dx}=\frac{1-(\frac{y}{x})^{3}}{(\frac{y}{x})^{2}}$

Let $\frac{y}{x}=V$ $\implies y=Vx$

$\frac{dy}{dx}=V+x\frac{dV}{dx}$

$V+x\frac{dV}{dx}=\frac{1-V^{3}}{V^{2}}$

$x\frac{dV}{dx}=\frac{1-V^{3}}{V^{2}}-V$

$x\frac{dV}{dx}=\frac{1-V^{3}-V^{3}}{V^{2}}$

$x\frac{dV}{dx}=\frac{1-2V^{3}}{V^{2}}$

$\frac{V^{2}}{1-2V^{3}}dV=\frac{1}{x}dx$

Integrating both sides

$\int \frac{V^{2}}{1-2V^{3}}dV=\int \frac{1}{x}dx$

Let 1-2V³=t

-6V²dV=dt

$V^{2}dV=-\frac{1}{6}dt$

$\frac{-1}{6}\int \frac{1}{t}dt=\int \frac{1}{x}dx$

$-\frac{1}{6}ln\ t=ln\ x+C$

$-\frac{1}{6}ln\ (1-2V^{3})=ln\ x+C$

$-\frac{1}{6}ln\ (1-2(\frac{y}{x})^{3})=ln\ x+C$

$-\frac{1}{6}ln[\frac{x^{3}-2y^{3}}{x^{3}}]=ln\ x+C$

(iii)

$(1+sin\ x)\frac{dy}{dx}+ycos\ x=tan\ x$

$\frac{dy}{dx}+\frac{ycos\ x}{(1+sin\ x)}=\frac{tan\ x}{(1+sin\ x)}$

This is a linear equation in y

Integrating factor ,I.F=$e^{\int{\frac{cos\ x}{1+sin\ x}}dx}$

$=e^{ln(1+sin\ x)}$

$=1+sin\ x$

Multiply through by I.F to get;

$\frac{d}{dx}[y(1+sin\ x)]=\frac{tan\ x}{1+sin\ x}(1+sin\ x)$

$\frac{d}{dx}[y(1+sin\ x)]={tan\ x}$

Integrating;

$y(1+sin\ x)=\int tan\ xdx$

$y(1+sin\ x)=ln|sec\ x|+C$

Our solution therefore is;

$y=\frac{ln|sex\ x|}{1+sin\ x}+\frac{C}{1+sin\ x}$

(iv)

$x\frac{dy}{dx}-5y=x^{7}$

$\frac{dy}{dx}-\frac{5y}{x}=x^{6}$

This is in the form $\frac{dy}{dx}+Py=Q$

$P=\frac{-5}{x},Q=x^{6}$

Integrating factor, I.F=$e^{\int Pdx}$

I.f=$e^{-5{\int \frac{1}{x}dx}}=e^{-5ln\ x}=e^{ln\ x^{-5}}=\frac{1}{x^{5}}$

$y*I.f=\int QI.f dx$

$y*\frac{1}{x^{5}}=\int x^{6}*\frac{1}{x^{5}}dx$

$\frac{y}{x^{5}}=\int xdx$

$\frac{y}{x^{5}}=\frac{x^{2}}{2}+C$

$y=\frac{x^{7}}{2}+Cx^{5}$