LinkedIn Insight UD-TRG-Maths2-Mumbai-Winter16-Q3-Ans - Grad Plus

UD-TRG-Maths2-Mumbai-Winter16-Q3-Ans

Qu. 3 (A) Evaluate [latex] I=\int_0^a\int_0^x\int_0^{x+y}e^{x+y+z}dx\;dy\;dz[/latex]

Sol. :- let, [latex] I=\int_0^a\int_0^x\int_0^{x+y}e^{x+y+z}dx\;dy\;dz[/latex]

Integrating w.r.f. z

[latex] I=\int_0^a\int_0^x\left[e^{x+y+z}\right]_0^{x+y}dy\;dx[/latex]

[latex] =\int_0^a\int_0^x\left[e^{x+y+\left(x+y\right)}-e^{x+y+o}\right]dy\;dx[/latex]

[latex] =\int_0^a\int_0^x\left[e^{2x+2y}-e^{x+y}\right]dy\;dx[/latex]

Integrating w.r.f. y

[latex] I=\int_0^a\left[\frac{e^{2x+2y}}2-e^{x+y}\right]_0^xdx[/latex]

[latex] I=\int_0^a\left(\frac{e^{2x+2y}}2-e^{x+2x}\right)-\left(\frac{e^{2x+0}}2-e^{x+0}\right)dx[/latex]

[latex] I=\int_0^a\left[\frac{e^{4x}}2-e^{2x}-\frac{e^{2x}}2+e^x\right]dx[/latex]

[latex] I=\int_0^a\left[\frac{e^{4x}}2-\frac{3e^{2x}}2+e^x\right]dx[/latex]

[latex] =\left[\frac12\times\frac{e^{4x}}4-\frac32\times\frac{e^{2x}}2+e^x\right]_0^a[/latex]

[latex] =\left[\frac18e^{4x}-\frac34e^{2x}+e^x\right]_0^a[/latex]

[latex] =\left[\frac18e^{4a}-\frac34e^{2a}+e^a\right]-\left[\frac18e^0-\frac34e^0+e^0\right][/latex]

[latex] =\left[\frac18e^{4a}-\frac34e^{2a}+e^a\right]-\left[\frac18-\frac34+1\right][/latex]

[latex] =\frac18e^{4a}-\frac34e^{2a}+e^a-\frac38[/latex]

[latex] =\frac18\left(e^{4a}-6e^{2a}+8e^a-3\right)[/latex]

[latex] \therefore\int_0^a\int_0^x\int_0^{x+y}e^{x+y+z}=\frac18\left(e^{4a}-6e^{2a}+8e^a-3\right)[/latex]

(C) Solve [latex] \left(2x+1\right)^2\frac{d^2y}{dx^2}-6\left(2x+1\right)\frac{dy}{dx}+16y=8\left(2x+1\right)^2[/latex]

sol. :- put [latex] \begin{array}{l}\left(2x+1\right)=e^z\\\therefore x=\log\;\left(2x+1\right)\\\therefore x=\frac{1+e^z}2\end{array}[/latex]

[latex] \begin{array}{l}\therefore\left(2x+1\right)\frac{dy}{dx}=\left(2D\right)y\\\end{array}[/latex]

[latex] \begin{array}{l}\left(2x-1\right)^2\frac{d^2y}{dx^2}=4D\left(D-1\right)y\;\;\&\;\;D=\frac d{d^z}\\\end{array}[/latex]

puting this value

We get,

[latex] \begin{array}{l}4D\left(D-1\right)y-6\left(2D\right)y+16y=8\left(e^z\right)^2\\\end{array}[/latex]

[latex] \begin{array}{l}\left(4D^2-4-12D+16\right)y=8\left(e^z\right)^2\\\end{array}[/latex]

[latex] \begin{array}{l}\left(4D^2-16D+16\right)y=8\left(e^z\right)^2\\\end{array}[/latex]

Dividing by 4

[latex] \begin{array}{l}\left(D^2-4D-4\right)y=2\left(e\right)^2\\\left(D^2-4D+4\right)y=2e^{2z}\\\end{array}[/latex]

Now,

put D = m

Its Auxiliary Equ is

[latex] \begin{array}{l}m^2-4m+4=0\\\therefore m=2,\;2\\\end{array}[/latex]

Therefore complimentary [latex] \begin{array}{l}f^h\left(c.f.\right)\;is\;\\yc\;=\;\left(c_1+zc_2\right)e^{2z}\end{array}[/latex]

Also,

Particular integral (P.I.) is yp = [latex] \frac1{f\left(D\right)}\times[/latex]

[latex] \begin{array}{l}\therefore yp=\frac1{\left(D-2\right)^2}2e^{2z}\\=2\times\frac z{2\left(D-2\right)}e^{2z}\end{array}[/latex]

[latex] \begin{array}{l}=z\times\frac z1e^{2z}\\yp=z^2e^{2z}\end{array}[/latex]

Therefore The complete solu. is y = yc = yp

[latex] \begin{array}{l}\therefore y=\left(c_1+zc_2\right)e^{2z}+z^2e^{2z}\\y=\left(c_1+zc_2+z^2\right)e^{2z}\end{array}[/latex]

[latex] y=c_1+\log\;\left(2x+1\right)c_2+\left(\log\;\left(2x+1\right)^2\right)e^{2\left(2x\right)}[/latex]

[latex] y=\left{c_1+\log\;\left(2x+1\right)c_2+\left[\log\left(2x+1\right)\right]^2\right}\;\left(2x+1\right)^2[/latex]

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