📐 Solutions to Mathematics-I Tutorial Sheet

First, we find \(\frac{dy}{dx}\). Using basic differentiation rules:

• The derivative of \(x = x^{1/2}\) is \(\frac{1}{2\sqrt{x}}\).
• The derivative of \(\frac{1}{\sqrt{x}} = x^{-1/2}\) is \(-\frac{1}{2}x^{-3/2} = -\frac{1}{2x^{3/2}}\).

Thus,

\[\frac{dy}{dx} = \frac{1}{2\sqrt{x}} - \frac{1}{2x^{3/2}}\]

Now compute \(2x\frac{dy}{dx} + y\):

\[2x\frac{dy}{dx} + y = 2x\left(\frac{1}{2\sqrt{x}} - \frac{1}{2x^{3/2}}\right) + \left(\sqrt{x} + \frac{1}{\sqrt{x}}\right)\]

Simplify term by term:

• \(2x \cdot \frac{1}{2\sqrt{x}} = \frac{2x}{2\sqrt{x}} = \sqrt{x}\)
• \(2x \cdot \left(-\frac{1}{2x^{3/2}}\right) = -\frac{2x}{2x^{3/2}} = -\frac{1}{\sqrt{x}}\)

So,

\[2x\frac{dy}{dx} + y = \sqrt{x} - \frac{1}{\sqrt{x}} + \sqrt{x} + \frac{1}{\sqrt{x}} = 2\sqrt{x}\]

Differentiate term by term:

• \(\frac{d}{dx}(\sin x) = \cos x\)
• \(\frac{d}{dx}(\cos x) = -\sin x\)

\[\frac{dy}{dx} = \cos x - \sin x\]

Now evaluate this at \(x = \frac{\pi}{3}\):

\[\frac{dy}{dx}\bigg|_{x=\pi/3} = \cos\frac{\pi}{3} - \sin\frac{\pi}{3} = \frac{1}{2} - \frac{\sqrt{3}}{2}\]

Differentiate term by term using \(\frac{d}{dx}(x^n) = nx^{n-1}\):

• \(\frac{d}{dx}(6x^5) = 30x^4\)
• \(\frac{d}{dx}(8x^4) = 32x^3\)
• \(\frac{d}{dx}(-2x^2) = -4x\)
• \(\frac{d}{dx}(5x) = 5\)
• \(\frac{d}{dx}(-9) = 0\)

\[\frac{dy}{dx} = 30x^4 + 32x^3 - 4x + 5\]

Now substitute \(x = 1\):

\[\frac{dy}{dx}\bigg|_{x=1} = 30(1)^4 + 32(1)^3 - 4(1) + 5 = 30 + 32 - 4 + 5 = 63\]

Differentiate each term:

• \(\frac{d}{dx}(x^2) = 2x\)
• \(\frac{d}{dx}\left(\frac{4}{x^2}\right) = \frac{d}{dx}(4x^{-2}) = -8x^{-3} = -\frac{8}{x^3}\)
• \(\frac{d}{dx}\left(-\frac{3}{2}\tan x\right) = -\frac{3}{2}\sec^2 x\), since \(\frac{d}{dx}(\tan x) = \sec^2 x\)
• \((6e)\) is a constant (where \(e \approx 2.718\)), so its derivative is 0

(i) \(y = x^{-3}\)

Using the power rule: \[\frac{dy}{dx} = -3x^{-4} = -\frac{3}{x^4}\]

(ii) \(y = e^x \sin 2x\)

Use the Product Rule \(\frac{d}{dx}(u \cdot v) = u'v + uv'\):

Let \(u = e^x\) and \(v = \sin 2x\)

• \(u' = e^x\)
• \(v' = 2\cos 2x\) (by chain rule)

\[\frac{dy}{dx} = e^x \cdot \sin 2x + e^x \cdot 2\cos 2x = e^x[\sin 2x + 2\cos 2x]\]

(iii) \(y = x^2 + \frac{4}{x^2} - \frac{3}{2}\tan x + 6e\)

This was solved in Problem 4:

\[\frac{dy}{dx} = 2x - \frac{8}{x^3} - \frac{3}{2}\sec^2 x\]

(iv) \(y = \sin(\sqrt{x})\)

Use chain rule. Let \(u = \sqrt{x} = x^{1/2}\), so \(y = \sin u\)

• \(\frac{du}{dx} = \frac{1}{2\sqrt{x}}\)
• \(\frac{dy}{du} = \cos u\)

\[\frac{dy}{dx} = \cos(\sqrt{x}) \cdot \frac{1}{2\sqrt{x}} = \frac{\cos(\sqrt{x})}{2\sqrt{x}}\]

(v) \(y = \sqrt{\sin x} = (\sin x)^{1/2}\)

Let \(u = \sin x\), so \(y = u^{1/2}\)

• \(\frac{du}{dx} = \cos x\)
• \(\frac{dy}{du} = \frac{1}{2\sqrt{u}} = \frac{1}{2\sqrt{\sin x}}\)

\[\frac{dy}{dx} = \frac{1}{2\sqrt{\sin x}} \cdot \cos x = \frac{\cos x}{2\sqrt{\sin x}}\]

(vi) \(y = \left(\sqrt{x} + \frac{1}{\sqrt{x}}\right)^2\)

Simplify first: \(\sqrt{x} + \frac{1}{\sqrt{x}} = \frac{x+1}{\sqrt{x}}\)

So \(y = \frac{(x+1)^2}{x}\)

Using quotient rule where \(u = (x+1)^2\) and \(v = x\):

• \(u' = 2(x+1)\)
• \(v' = 1\)

\[\frac{dy}{dx} = \frac{2(x+1) \cdot x - (x+1)^2 \cdot 1}{x^2} = \frac{2x^2 + 2x - x^2 - 2x - 1}{x^2} = \frac{x^2 - 1}{x^2}\]

(vii) \(y = \sin(\sqrt{\sin x} + \cos x)\)

Let \(u = \sqrt{\sin x} + \cos x\), then \(y = \sin(u)\)

\[\frac{du}{dx} = \frac{\cos x}{2\sqrt{\sin x}} - \sin x\]

\[\frac{dy}{du} = \cos u = \cos(\sqrt{\sin x} + \cos x)\]

\[\frac{dy}{dx} = \cos(\sqrt{\sin x} + \cos x) \cdot \left(\frac{\cos x}{2\sqrt{\sin x}} - \sin x\right)\]

(viii) \(y = \sqrt{x^3} = x^{3/2}\)

\[\frac{dy}{dx} = \frac{3}{2}x^{1/2} = \frac{3}{2}\sqrt{x}\]

(ix) \(y = \frac{1}{\sqrt{x^3}} = x^{-3/2}\)

\[\frac{dy}{dx} = -\frac{3}{2}x^{-5/2} = -\frac{3}{2x^{5/2}} = -\frac{3}{2\sqrt{x^5}}\]

(x) \(y = 3x^2 + 2x + 5\sqrt{x}\)

\[\frac{dy}{dx} = 6x + 2 + \frac{5}{2\sqrt{x}}\]

(xi) \(y = \sin(x^2)\)

Let \(u = x^2\), then \(y = \sin u\)

\[\frac{dy}{dx} = \cos(x^2) \cdot 2x = 2x\cos(x^2)\]

(xii) \(y = \sin 2x\)

\[\frac{dy}{dx} = 2\cos 2x\]

(xiii) \(y = \cos(x^2 + 1)\)

\[\frac{dy}{dx} = -\sin(x^2 + 1) \cdot 2x = -2x\sin(x^2 + 1)\]

(xiv) \(y = e^{\frac{x}{1+\sin x}}\)

Let \(u = \frac{x}{1+\sin x}\), then \(y = e^u\)

Using quotient rule for \(u\):

\[u' = \frac{(1+\sin x) \cdot 1 - x \cdot \cos x}{(1+\sin x)^2} = \frac{1 + \sin x - x\cos x}{(1+\sin x)^2}\]

\[\frac{dy}{dx} = e^{\frac{x}{1+\sin x}} \cdot \frac{1 + \sin x - x\cos x}{(1+\sin x)^2}\]

(xv) \(y = \frac{1 + \tan x}{1 - \tan x}\)

Using quotient rule where \(p = 1 - \tan x\) and \(q = 1 + \tan x\):

• \(p' = -\sec^2 x\)
• \(q' = \sec^2 x\)

\[\frac{dy}{dx} = \frac{(-\sec^2 x)(1+\tan x) - (1-\tan x)(\sec^2 x)}{(1+\tan x)^2}\]

Simplifying the numerator:

\[= \frac{-\sec^2 x - \sec^2 x\tan x - \sec^2 x + \sec^2 x\tan x}{(1+\tan x)^2} = \frac{-2\sec^2 x}{(1+\tan x)^2}\]

Let \(F(x,y,z) = x^3 + y^3 + z^3 - 8xyz\), so \(u = \log[F(x,y,z)]\).

Using the chain rule: \(\frac{\partial u}{\partial x} = \frac{1}{F(x,y,z)} \cdot \frac{\partial F}{\partial x}\)

With respect to \(x\):

\[\frac{\partial F}{\partial x} = 3x^2 - 8yz\]

\[\frac{\partial u}{\partial x} = \frac{3x^2 - 8yz}{x^3 + y^3 + z^3 - 8xyz}\]

With respect to \(y\):

\[\frac{\partial F}{\partial y} = 3y^2 - 8xz\]

\[\frac{\partial u}{\partial y} = \frac{3y^2 - 8xz}{x^3 + y^3 + z^3 - 8xyz}\]

With respect to \(z\):

\[\frac{\partial F}{\partial z} = 3z^2 - 8xy\]

\[\frac{\partial u}{\partial z} = \frac{3z^2 - 8xy}{x^3 + y^3 + z^3 - 8xyz}\]

Let \(G(x,y,z) = x^4 + y^3 + z^2 - 12xyz\), so \(u = \log[G(x,y,z)]\).

Compute partial derivatives of \(G\):

• \(\frac{\partial G}{\partial x} = 4x^3 - 12yz\)
• \(\frac{\partial G}{\partial y} = 3y^2 - 12xz\)
• \(\frac{\partial G}{\partial z} = 2z - 12xy\)

Let \(H(x,y) = x^3 + y^3 - xy\), so \(u = \sin[H(x,y)]\).

Using chain rule:

\[\frac{\partial u}{\partial x} = \cos[H(x,y)] \cdot \frac{\partial H}{\partial x}\]

\[\frac{\partial u}{\partial y} = \cos[H(x,y)] \cdot \frac{\partial H}{\partial y}\]

Compute partial derivatives:

• \(\frac{\partial H}{\partial x} = 3x^2 - y\)
• \(\frac{\partial H}{\partial y} = 3y^2 - x\)

Here \(u\) is a function of the ratio \(v = \frac{y}{x}\). So \(u(x,y) = f(v)\) where \(v = \frac{y}{x}\).

Using chain rule:

\[\frac{\partial u}{\partial x} = f'(v) \cdot \frac{\partial v}{\partial x}\]

\[\frac{\partial u}{\partial y} = f'(v) \cdot \frac{\partial v}{\partial y}\]

Computing partials of \(v = \frac{y}{x} = yx^{-1}\):

• \(\frac{\partial v}{\partial x} = y \cdot (-x^{-2}) = -\frac{y}{x^2}\)
• \(\frac{\partial v}{\partial y} = \frac{1}{x}\)

Therefore:

\[\frac{\partial u}{\partial x} = f'(v) \left(-\frac{y}{x^2}\right)\]

\[\frac{\partial u}{\partial y} = f'(v) \left(\frac{1}{x}\right)\]

Now form \(xu_x + yu_y\):

\[x \cdot f'(v)\left(-\frac{y}{x^2}\right) + y \cdot f'(v)\left(\frac{1}{x}\right)\]

\[= -\frac{f'(v)y}{x} + \frac{f'(v)y}{x} = 0\]

Compute partial derivatives:

• \(\frac{\partial u}{\partial x} = 2x\)
• \(\frac{\partial u}{\partial y} = 2y\)

Now form \(xu_x + yu_y\):

\[x(2x) + y(2y) = 2x^2 + 2y^2 = 2(x^2 + y^2) = 2u\]

Using total derivative formula:

\[\frac{du}{dt} = \frac{\partial u}{\partial x}\frac{dx}{dt} + \frac{\partial u}{\partial y}\frac{dy}{dt}\]

Find partial derivatives of \(u = x\ln y\):

• \(\frac{\partial u}{\partial x} = \ln y\)
• \(\frac{\partial u}{\partial y} = \frac{x}{y}\)

Find derivatives with respect to \(t\):

• \(\frac{dx}{dt} = 2t\) (since \(x = t^2\))
• \(\frac{dy}{dt} = e^t\) (since \(y = e^t\))

Substitute into total derivative:

\[\frac{du}{dt} = (\ln y)(2t) + \left(\frac{x}{y}\right)(e^t)\]

Since \(y = e^t\), we have \(\ln y = t\) and \(\frac{1}{y} = e^{-t}\).

Also \(x = t^2\).

\[\frac{du}{dt} = (t)(2t) + (t^2 \cdot e^{-t})(e^t) = 2t^2 + t^2 = 3t^2\]

Using total derivative formula:

\[\frac{du}{dt} = \frac{\partial u}{\partial x}\frac{dx}{dt} + \frac{\partial u}{\partial y}\frac{dy}{dt} + \frac{\partial u}{\partial z}\frac{dz}{dt}\]

Compute partial derivatives of \(u = xy + z\):

• \(\frac{\partial u}{\partial x} = y\)
• \(\frac{\partial u}{\partial y} = x\)
• \(\frac{\partial u}{\partial z} = 1\)

Find derivatives:

• \(\frac{dx}{dt} = -\sin t\)
• \(\frac{dy}{dt} = \cos t\)
• \(\frac{dz}{dt} = 1\)

Substitute:

\[\frac{du}{dt} = (y)(-\sin t) + (x)(\cos t) + (1)(1)\]

\[= (\sin t)(-\sin t) + (\cos t)(\cos t) + 1\]

\[= -\sin^2 t + \cos^2 t + 1 = \cos 2t + 1\]

Let \(I = \displaystyle \int_0^{\pi/2} \frac{\sin x}{\sin x + \cos x}\,dx\)

We use the property that for integrals over \([0, \pi/2]\), if we substitute \(x' = \frac{\pi}{2} - x\), we get:

\[I' = \int_0^{\pi/2} \frac{\cos x}{\sin x + \cos x}\,dx\]

By symmetry, \(I = I'\). Now add them:

\[I + I' = \int_0^{\pi/2} \frac{\sin x + \cos x}{\sin x + \cos x}\,dx = \int_0^{\pi/2} 1\,dx = \frac{\pi}{2}\]

Since \(I = I'\), we have \(2I = \frac{\pi}{2}\), thus:

Let \(J = \displaystyle \int_0^{\pi/2} \frac{\sqrt{\sin x}}{\sqrt{\sin x} + \sqrt{\cos x}}\,dx\)

Define \(J' = \displaystyle \int_0^{\pi/2} \frac{\sqrt{\cos x}}{\sqrt{\sin x} + \sqrt{\cos x}}\,dx\)

Adding:

\[J + J' = \int_0^{\pi/2} \frac{\sqrt{\sin x} + \sqrt{\cos x}}{\sqrt{\sin x} + \sqrt{\cos x}}\,dx = \int_0^{\pi/2} 1\,dx = \frac{\pi}{2}\]

By symmetry (substituting \(x' = \pi/2 - x\)), we get \(J = J'\).

Therefore, \(2J = \frac{\pi}{2}\), so:

Observe that \(\sin^2 x + \cos^2 x = 1\) for all \(x\).

So the integrand simplifies:

\[\frac{\sin^2 x}{\sin^2 x + \cos^2 x} = \frac{\sin^2 x}{1} = \sin^2 x\]

We need to evaluate \(\displaystyle \int_0^{\pi/2} \sin^2 x\,dx\)

Using the identity \(\sin^2 x = \frac{1 - \cos 2x}{2}\):

\[\int_0^{\pi/2} \sin^2 x\,dx = \int_0^{\pi/2} \frac{1 - \cos 2x}{2}\,dx\]

\[= \frac{1}{2}\int_0^{\pi/2} 1\,dx - \frac{1}{2}\int_0^{\pi/2} \cos 2x\,dx\]

Compute each integral:

• \(\displaystyle \int_0^{\pi/2} 1\,dx = \frac{\pi}{2}\)
• \(\displaystyle \int_0^{\pi/2} \cos 2x\,dx = \left[\frac{\sin 2x}{2}\right]_0^{\pi/2} = \frac{\sin \pi - \sin 0}{2} = 0\)

Therefore:

\[\int_0^{\pi/2} \sin^2 x\,dx = \frac{1}{2} \cdot \frac{\pi}{2} - \frac{1}{2} \cdot 0 = \frac{\pi}{4}\]

The parabola \(8x = y^2\) can be written as \(x = \frac{y^2}{8}\). It opens to the right with vertex at origin.

Find intersection with \(x = 4\):

\[\frac{y^2}{8} = 4 \implies y^2 = 32 \implies y = \pm 4\sqrt{2}\]

The region is bounded between \(y = -4\sqrt{2}\) and \(y = 4\sqrt{2}\).

For a horizontal strip at height \(y\), the width is:

\[\text{width} = 4 - \frac{y^2}{8}\]

Area:

\[A = \int_{-4\sqrt{2}}^{4\sqrt{2}} \left(4 - \frac{y^2}{8}\right)dy\]

Using symmetry:

\[A = 2\int_0^{4\sqrt{2}} \left(4 - \frac{y^2}{8}\right)dy = 2\left[4y - \frac{y^3}{24}\right]_0^{4\sqrt{2}}\]

At \(y = 4\sqrt{2}\):

• \(4y = 16\sqrt{2}\)
• \(\frac{y^3}{24} = \frac{(4\sqrt{2})^3}{24} = \frac{128\sqrt{2}}{24} = \frac{16\sqrt{2}}{3}\)

\[A = 2\left(16\sqrt{2} - \frac{16\sqrt{2}}{3}\right) = 2 \cdot \frac{32\sqrt{2}}{3} = \frac{64\sqrt{2}}{3}\]

The parabola is \(x = \frac{y^2}{4a}\). Find intersection with \(x = 1\):

\[\frac{y^2}{4a} = 1 \implies y^2 = 4a \implies y = \pm 2\sqrt{a}\]

The width of horizontal strip at height \(y\) is:

\[\text{width} = 1 - \frac{y^2}{4a}\]

Area:

\[A = \int_{-2\sqrt{a}}^{2\sqrt{a}} \left(1 - \frac{y^2}{4a}\right)dy = 2\int_0^{2\sqrt{a}} \left(1 - \frac{y^2}{4a}\right)dy\]

\[A = 2\left[y - \frac{y^3}{12a}\right]_0^{2\sqrt{a}}\]

At \(y = 2\sqrt{a}\):

• \(y = 2\sqrt{a}\)
• \(\frac{y^3}{12a} = \frac{8a^{3/2}}{12a} = \frac{2\sqrt{a}}{3}\)

\[A = 2\left(2\sqrt{a} - \frac{2\sqrt{a}}{3}\right) = 2 \cdot \frac{4\sqrt{a}}{3} = \frac{8\sqrt{a}}{3}\]

Find intersection: \(y^2 = 2 \implies y = \pm\sqrt{2}\)

Width of horizontal strip: \(2 - y^2\)

\[A = \int_{-\sqrt{2}}^{\sqrt{2}} (2 - y^2)dy = 2\int_0^{\sqrt{2}} (2 - y^2)dy\]

\[= 2\left[2y - \frac{y^3}{3}\right]_0^{\sqrt{2}}\]

\[= 2\left(2\sqrt{2} - \frac{2\sqrt{2}}{3}\right) = 2 \cdot \frac{4\sqrt{2}}{3} = \frac{8\sqrt{2}}{3}\]

First integrate with respect to \(y\):

\[\int_{-1}^{1} (x + y)dy = \int_{-1}^{1} x\,dy + \int_{-1}^{1} y\,dy\]

• \(\int_{-1}^{1} x\,dy = x[y]_{-1}^{1} = x(2) = 2x\)
• \(\int_{-1}^{1} y\,dy = \left[\frac{y^2}{2}\right]_{-1}^{1} = 0\)

Inner integral = \(2x\)

Now integrate with respect to \(x\):

\[\int_0^1 2x\,dx = 2\left[\frac{x^2}{2}\right]_0^1 = 2 \cdot \frac{1}{2} = 1\]

Step 1: Integrate with respect to \(z\):

\[I_1(x,y) = \int_1^2 (2x^2 + y + z^3)dz = 2x^2[z]_1^2 + y[z]_1^2 + \left[\frac{z^4}{4}\right]_1^2\]

\[= 2x^2(1) + y(1) + \frac{15}{4} = 2x^2 + y + \frac{15}{4}\]

Step 2: Integrate with respect to \(y\):

\[I_2(x) = \int_0^3 \left(2x^2 + y + \frac{15}{4}\right)dy = 2x^2[y]_0^3 + \left[\frac{y^2}{2}\right]_0^3 + \frac{15}{4}[y]_0^3\]

\[= 6x^2 + \frac{9}{2} + \frac{45}{4} = 6x^2 + \frac{63}{4}\]

Step 3: Integrate with respect to \(x\):

\[I = \int_1^4 \left(6x^2 + \frac{63}{4}\right)dx = 6\left[\frac{x^3}{3}\right]_1^4 + \frac{63}{4}[x]_1^4\]

\[= 2(64-1) + \frac{63}{4}(3) = 126 + \frac{189}{4} = \frac{693}{4}\]

Step 1: Integrate with respect to \(z\):

\[J_1(x,y) = \int_1^2 (3x^2 - y + z^3)dz = 3x^2 - y + \frac{15}{4}\]

Step 2: Integrate with respect to \(y\):

\[J_2(x) = \int_0^3 \left(3x^2 - y + \frac{15}{4}\right)dy = 9x^2 - \frac{9}{2} + \frac{45}{4} = 9x^2 + \frac{27}{4}\]

Step 3: Integrate with respect to \(x\):

\[J = \int_1^4 \left(9x^2 + \frac{27}{4}\right)dx = 3(64-1) + \frac{27}{4}(3) = 189 + \frac{81}{4} = \frac{837}{4}\]

Rewrite in standard form by dividing by \(x\):

\[\frac{dy}{dx} - \frac{y}{x} = 2x^2\]

This is a first-order linear ODE: \(\frac{dy}{dx} + P(x)y = Q(x)\) where \(P(x) = -\frac{1}{x}\) and \(Q(x) = 2x^2\).

Integrating factor:

\[\mu(x) = e^{\int P(x)dx} = e^{\int -\frac{1}{x}dx} = e^{-\ln|x|} = x^{-1}\]

Multiply the equation by \(\mu(x) = x^{-1}\):

\[x^{-1}\frac{dy}{dx} - \frac{y}{x^2} = 2x\]

The left side is \(\frac{d}{dx}(y \cdot x^{-1})\), so:

\[\frac{d}{dx}\left(\frac{y}{x}\right) = 2x\]

Integrate both sides:

\[\frac{y}{x} = \int 2x\,dx = x^2 + C\]

Therefore:

Divide by \(2x\):

\[\frac{dy}{dx} + \frac{y}{2x} = 2x^2\]

Here \(P(x) = \frac{1}{2x}\) and \(Q(x) = 2x^2\).

Integrating factor:

\[\mu(x) = e^{\int \frac{1}{2x}dx} = e^{\frac{1}{2}\ln|x|} = |x|^{1/2} = \sqrt{x}\]

Multiply by \(\sqrt{x}\):

\[\sqrt{x}\frac{dy}{dx} + \frac{y}{2\sqrt{x}} = 2x^{5/2}\]

This is \(\frac{d}{dx}(\sqrt{x}\,y) = 2x^{5/2}\)

Integrate:

\[\sqrt{x}\,y = \int 2x^{5/2}dx = \frac{4}{7}x^{7/2} + C\]

Therefore:

This is separable. Separate variables:

\[\frac{1}{y^2}dy = -\frac{1}{x^2}dx\]

Integrate both sides:

\[\int y^{-2}dy = \int -x^{-2}dx\]

\[-\frac{1}{y} = \frac{1}{x} + C\]

Solve for \(y\):

\[\frac{1}{y} = C - \frac{1}{x} = \frac{Cx - 1}{x}\]

Separate variables:

\[\frac{1}{y}dy = \frac{7x}{1+x^2}dx\]

Integrate both sides:

\[\ln|y| = \int \frac{7x}{1+x^2}dx = \frac{7}{2}\ln|1+x^2| + C\]

Exponentiate:

\[|y| = e^{\frac{7}{2}\ln(1+x^2) + C} = A(1+x^2)^{7/2}\]

Separate variables:

\[\frac{1}{y}dy = \frac{3x^2}{1+x^3}dx\]

Integrate:

\[\ln|y| = \int \frac{3x^2}{1+x^3}dx = \ln|1+x^3| + C\]

Exponentiate:

\[|y| = e^C|1+x^3|\]

Matrix multiplication (row by column):

\[AB = \begin{pmatrix} 1 & 3 \\ 0 & -1 \end{pmatrix} \begin{pmatrix} 1 & 3 \\ 0 & -1 \end{pmatrix}\]

Calculate each element:

• Top-left: \(1 \cdot 1 + 3 \cdot 0 = 1\)
• Top-right: \(1 \cdot 3 + 3 \cdot (-1) = 0\)
• Bottom-left: \(0 \cdot 1 + (-1) \cdot 0 = 0\)
• Bottom-right: \(0 \cdot 3 + (-1) \cdot (-1) = 1\)

Since \(B\) is the identity matrix:

\[AB = A \cdot I = A\]

From equation (1): \(x = 1 + y - z\)

Substitute into equations (2) and (3):

• From (2): \(4(1 + y - z) + 3y - z = 6\) gives \(7y - 5z = 2\)
• From (3): \(3(1 + y - z) + 5y + 3z = 4\) gives \(8y = 1\), so \(y = \frac{1}{8}\)

From \(7y - 5z = 2\) and \(y = \frac{1}{8}\):

\[7 \cdot \frac{1}{8} - 5z = 2 \implies -5z = 2 - \frac{7}{8} = \frac{9}{8}\]

\[z = -\frac{9}{40}\]

From equation (1):

\[x = 1 + \frac{1}{8} - \left(-\frac{9}{40}\right) = 1 + \frac{1}{8} + \frac{9}{40} = \frac{27}{20}\]

From equation (1): \(x = 9 - y - z\)

Substitute into (2) and (3):

• From (2): \(5y - 2z = 5\)
• From (3): \(y + 2z = 13\)

From \(y + 2z = 13\): \(y = 13 - 2z\)

Substitute into \(5y - 2z = 5\):

\[5(13 - 2z) - 2z = 5 \implies 65 - 12z = 5 \implies z = 5\]

Then \(y = 13 - 2(5) = 3\)

And \(x = 9 - 3 - 5 = 1\)

From equation (1): \(x = 4 + y - z\)

Substitute into (2) and (3):

• From (2): \(3y - 5z = -8\)
• From (3): \(2y = -2\), so \(y = -1\)

From \(3y - 5z = -8\) with \(y = -1\):

\[3(-1) - 5z = -8 \implies -5z = -5 \implies z = 1\]

Then \(x = 4 + (-1) - 1 = 2\)

From equation (1): \(x = -4 - 2y + 3z\)

Substitute into (2) and (3):

• From (2): \(y - 8z = -10\)
• From (3): \(-9y + 5z = 23\)

From \(y - 8z = -10\): \(y = -10 + 8z\)

Substitute into \(-9y + 5z = 23\):

\[-9(-10 + 8z) + 5z = 23 \implies 90 - 67z = 23 \implies z = 1\]

Then \(y = -10 + 8(1) = -2\)

And \(x = -4 - 2(-2) + 3(1) = 3\)