JEE Main 2022 — Mathematics Calculus

Given, $(x-{y}^{2})dx+y(5x+{y}^{2})dy=0$(x−y2)dx+y(5x+y2)dy=0

After rearranging we get,

$$\frac{dy}{dx}=\frac{{y}^{2}-x}{y(5x+{y}^{2})}$$dxdy​=y(5x+y2)y2−x​

Now, let ${y}^{2}=v\Rightarrow 2y\frac{dy}{dx}=\frac{dv}{dx}$y2=v⇒2ydxdy​=dxdv​

$$\frac{1}{2y}\frac{dv}{dx}=\frac{v-x}{y(5x+v)}$$2y1​dxdv​=y(5x+v)v−x​

$$\frac{dv}{dx}=2(\frac{v-x}{5x+v})$$dxdv​=2(5x+vv−x​)

Now let $v=kx$v=kx

$$\Rightarrow k+x\frac{dk}{dx}=2(\frac{kx-x}{5x+kx})$$⇒k+xdxdk​=2(5x+kxkx−x​)

$$\Rightarrow x\frac{dk}{dx}=-\frac{({k}^{2}+3k+2)}{k+5}$$⇒xdxdk​=−k+5(k2+3k+2)​

$$\Rightarrow \int \frac{(5+k)}{(k+1)(k+2)}dk=\int -\frac{dx}{x}$$⇒∫(k+1)(k+2)(5+k)​dk=∫−xdx​

$$\Rightarrow \int (\frac{4}{k+1}-\frac{3}{k+2})dk=-\int \frac{dx}{x}$$⇒∫(k+14​−k+23​)dk=−∫xdx​

$$\Rightarrow 4\mathrm{ln}(k+1)-3\mathrm{ln}(k+2)=-\mathrm{ln}x+\mathrm{ln}c$$⇒4ln(k+1)−3ln(k+2)=−lnx+lnc

$$\Rightarrow \frac{{(k+1)}^{4}}{{(k+2)}^{3}}=\frac{c}{x}$$⇒(k+2)3(k+1)4​=xc​

$$\Rightarrow \frac{{(\frac{v}{x}+1)}^{4}}{{(\frac{v}{x}+2)}^{3}}=\frac{c}{x}$$⇒(xv​+2)3(xv​+1)4​=xc​

$$\Rightarrow c{({y}^{2}+2x)}^{3}={({y}^{2}+x)}^{4}$$⇒c(y2+2x)3=(y2+x)4