If the curve y=y(x) is the solution of the differential equation 2(x^2+x^5/4)dy-y(x+x^1/4)dx=2x^9/4dx,x>0 which passes through the point (1,1- 4 3…

Given differential equation is $2({x}^{2}+{x}^{5/4})dy-y(x+{x}^{1/4})dx=2{x}^{9/4}dx$2(x2+x5/4)dy−y(x+x1/4)dx=2x9/4dx

$$\Rightarrow 2({x}^{2}+{x}^{5/4})\frac{dy}{dx}-y(x+{x}^{1/4})=2{x}^{9/4}$$⇒2(x2+x5/4)dxdy​−y(x+x1/4)=2x9/4

$$\Rightarrow \frac{dy}{dx}-\frac{y(x+{x}^{1/4})}{2({x}^{2}+{x}^{5/4})}=\frac{2{x}^{9/4}}{2({x}^{2}+{x}^{5/4})}$$⇒dxdy​−2(x2+x5/4)y(x+x1/4)​=2(x2+x5/4)2x9/4​

$$\Rightarrow \frac{dy}{dx}-\frac{y(x+{x}^{1/4})}{2x(x+{x}^{1/4})}=\frac{2{x}^{9/4}}{2{x}^{5/4}({x}^{3/4}+1)}$$⇒dxdy​−2x(x+x1/4)y(x+x1/4)​=2x5/4(x3/4+1)2x9/4​

$$\Rightarrow \frac{dy}{dx}-\frac{y}{2x}=\frac{x}{({x}^{3/4}+1)}$$⇒dxdy​−2xy​=(x3/4+1)x​

This is a linear differential equation of the type $\frac{dy}{dx}+Py=Q,$dxdy​+Py=Q, where $P=-\frac{1}{2x}$P=−2x1​ and $Q=\frac{x}{{x}^{3/4}+1}$Q=x3/4+1x​

Now, we have integrating factor $\displaystyle I.F.={e}^{\int Pdx}$I.F.=e∫Pdx

$$={e}^{-\int \frac{dx}{2x}}={e}^{-\frac{1}{2}{\mathrm{log}}_{e}x}$$=e−∫2xdx​=e−21​loge​x

$$={e}^{{\mathrm{log}}_{e}{x}^{-\frac{1}{2}}}={x}^{-\frac{1}{2}}$$=eloge​x−21​=x−21​

Now, the solution of the linear differential equation is $\displaystyle y\times (I.F.)=\int Q\times (I.F.)dx+C$y×(I.F.)=∫Q×(I.F.)dx+C

$$\Rightarrow y\cdot {x}^{-1/2}=\int \frac{x\cdot {x}^{-1/2}}{({x}^{3/4}+1)}dx+C$$⇒y⋅x−1/2=∫(x3/4+1)x⋅x−1/2​dx+C

$$\Rightarrow y\cdot {x}^{-1/2}=\int \frac{{x}^{1/2}}{({x}^{3/4}+1)}dx+C$$⇒y⋅x−1/2=∫(x3/4+1)x1/2​dx+C

Put $x={t}^{4}\Rightarrow dx=4{t}^{3}dt$x=t4⇒dx=4t3dt

$$\Rightarrow y\cdot {x}^{-1/2}=\int \frac{{t}^{2}\cdot 4{t}^{3}dt}{({t}^{3}+1)}$$⇒y⋅x−1/2=∫(t3+1)t2⋅4t3dt​

$$\Rightarrow y\cdot {x}^{-1/2}=4\int \frac{{t}^{2}({t}^{3}+1-1)}{({t}^{3}+1)}dt$$⇒y⋅x−1/2=4∫(t3+1)t2(t3+1−1)​dt

$$\Rightarrow y\cdot {x}^{-1/2}=4\int {t}^{2}dt-4\int \frac{{t}^{2}}{{t}^{3}+1}dt+C$$⇒y⋅x−1/2=4∫t2dt−4∫t3+1t2​dt+C

Let, ${t}^{3}+1=u\Rightarrow 3{t}^{2}dt=du$t3+1=u⇒3t2dt=du

$$\Rightarrow y\cdot {x}^{-1/2}=4\int {t}^{2}dt-\frac{4}{3}\int \frac{du}{u}+C$$⇒y⋅x−1/2=4∫t2dt−34​∫udu​+C

$$\Rightarrow y\cdot {x}^{-1/2}=\frac{4{t}^{3}}{3}-\frac{4}{3}{\mathrm{log}}_{e}|u|+C$$⇒y⋅x−1/2=34t3​−34​loge​∣u∣+C

$$\Rightarrow y\cdot {x}^{-1/2}=\frac{4{t}^{3}}{3}-\frac{4}{3}{\mathrm{log}}_{e}({t}^{3}+1)+C$$⇒y⋅x−1/2=34t3​−34​loge​(t3+1)+C

$$\Rightarrow y\cdot {x}^{-1/2}=\frac{4{x}^{3/4}}{3}-\frac{4}{3}{\mathrm{log}}_{e}({x}^{3/4}+1)+C$$⇒y⋅x−1/2=34x3/4​−34​loge​(x3/4+1)+C

Now, the curve passes through $(1,1-\frac{4}{3}{\mathrm{log}}_{e}2)$(1,1−34​loge​2)

$$\Rightarrow 1-\frac{4}{3}{\mathrm{log}}_{e}2=\frac{4}{3}-\frac{4}{3}{\mathrm{log}}_{e}2+C$$⇒1−34​loge​2=34​−34​loge​2+C

$$\Rightarrow C=-\frac{1}{3}$$⇒C=−31​

$$\Rightarrow y\cdot {x}^{-1/2}=\frac{4}{3}{x}^{3/4}-\frac{4}{3}{\mathrm{log}}_{e}({x}^{3/4}+1)-\frac{1}{3}$$⇒y⋅x−1/2=34​x3/4−34​loge​(x3/4+1)−31​

$$\Rightarrow y=\frac{4}{3}{x}^{5/4}-\frac{4}{3}\sqrt{x}{\mathrm{log}}_{e}({x}^{3/4}+1)-\frac{\sqrt{x}}{3}$$⇒y=34​x5/4−34​x​loge​(x3/4+1)−3x​​

Hence, at $x=16,$x=16, we get

$$y(16)=\frac{4}{3}\times 32-\frac{4}{3}\times 4\times {\mathrm{log}}_{e}9-\frac{4}{3}$$y(16)=34​×32−34​×4×loge​9−34​

$$\Rightarrow y(16)=\frac{124}{3}-\frac{32}{3}{\mathrm{log}}_{e}3$$⇒y(16)=3124​−332​loge​3

$$\Rightarrow y(16)=4(\frac{31}{3}-\frac{8}{3}{\mathrm{log}}_{e}3).$$⇒y(16)=4(331​−38​loge​3).